Motion in One Dimension - Questions & Answers
Step By Step Explained
EXERCISE-2(C)(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).
1. When a body starts from rest, the equation of motion takes the form:
(a) v = u
(b) v = at
(c) v = (1/2) at²
(d) S = ut + (1/2) at²
Step 1: The standard first equation of motion is v = u + at.
Step 2: "Starts from rest" means the initial velocity (u) is zero.
Step 3: Substitute u = 0 into the equation.
Step 4: The equation becomes v = 0 + at, which simplifies to v = at.
Answer: (b) v = at
2. If a body is moving with a uniform retardation, then its acceleration will be :
(a) Positive
(b) Negative
(c) Zero
(d) Cannot say
Step 1: Retardation implies that the velocity of the moving body is decreasing over time.
Step 2: Acceleration is defined as the rate of change of velocity.
Step 3: When final velocity is less than initial velocity, the change in velocity is negative.
Step 4: Therefore, retardation is equivalent to a negative acceleration.
Answer: (b) Negative
3. A car acquires a velocity of 54 ms⁻¹ in 20 s starting from rest, then its acceleration is :
(a) 5.4 ms⁻²
(b) 2.7 ms⁻²
(c) 7.2 ms⁻²
(d) 2.0 ms⁻²
Step 1: Identify the given values from the question.
Step 2: The car starts from rest, so initial velocity (u) = 0 m s⁻¹.
Step 3: The acquired final velocity (v) = 54 m s⁻¹.
Step 4: The time taken (t) = 20 s.
Step 5: Use the formula for acceleration: a = (v - u) / t.
Step 6: Substitute the values: a = (54 - 0) / 20.
Step 7: Calculate the result: a = 54 / 20 = 2.7 m s⁻².
Answer: (b) 2.7 ms⁻²
4. A particle starts to move in a straight line from a point with velocity 10 ms⁻¹ and acceleration -2.0 ms⁻². Its position at t = 5s will be :
(a) 5 m
(b) 10 m
(c) 20 m
(d) 25 m
Step 1: Identify the given values.
Step 2: Initial velocity (u) = 10 m s⁻¹.
Step 3: Acceleration (a) = -2.0 m s⁻².
Step 4: Time (t) = 5 s.
Step 5: Use the second equation of motion for displacement: S = ut + (1/2)at².
Step 6: Substitute the values: S = (10 × 5) + (1/2) × (-2.0) × (5)².
Step 7: Calculate the first part: 10 × 5 = 50.
Step 8: Calculate the second part: (1/2) × (-2.0) × 25 = -25.
Step 9: Add them together: S = 50 - 25 = 25 m.
Answer: (d) 25 m
5. A body initially at rest, starts moving with a constant acceleration of 0.5 ms⁻² and travels a distance 25 m, then its final velocity is :
(a) 5 ms⁻¹
(b) 20 ms⁻¹
(c) 15 ms⁻¹
(d) -15 ms⁻¹
Step 1: Identify the given values.
Step 2: Initial velocity (u) = 0 (since it is at rest).
Step 3: Constant acceleration (a) = 0.5 m s⁻².
Step 4: Distance travelled (S) = 25 m.
Step 5: Use the third equation of motion: v² = u² + 2aS.
Step 6: Substitute the values into the formula: v² = (0)² + 2 × 0.5 × 25.
Step 7: Perform the multiplication: 2 × 0.5 = 1, and 1 × 25 = 25.
Step 8: This leaves v² = 25.
Step 9: Take the square root of both sides to find v: v = 5 m s⁻¹.
Answer: (a) 5 ms⁻¹
6. A car starting from rest accelerates uniformly to acquire a speed 20 km h⁻¹ in 30 min. The distance travelled by car in this time interval will be :
(a) 600 km
(b) 5 km
(c) 6 km
(d) 10 km
Step 1: Identify the given values.
Step 2: Initial velocity (u) = 0 km h⁻¹.
Step 3: Final velocity (v) = 20 km h⁻¹.
Step 4: Time (t) = 30 min. Convert minutes to hours: t = 30 / 60 = 0.5 h.
Step 5: Since acceleration is uniform, we can use the Average Velocity formula: Average Velocity = (u + v) / 2.
Step 6: Calculate average velocity: (0 + 20) / 2 = 10 km h⁻¹.
Step 7: Use the formula: Distance (S) = Average Velocity × Time.
Step 8: Substitute the values: S = 10 × 0.5.
Step 9: Calculate the distance: S = 5 km.
Answer: (b) 5 km
7. A car starting from rest moves on a straight path for time t = 0 to t = T with a uniform acceleration a and then stops with a uniform retardation. The average speed of the car will be :
(a) aT/4
(b) 3aT/2
(c) aT/2
(d) aT
Step 1: The car starts from rest (u = 0) and accelerates for time T with uniform acceleration a.
Step 2: The maximum velocity (V) it reaches at time T is V = u + aT = 0 + aT = aT.
Step 3: It then undergoes uniform retardation and eventually stops, so its final velocity is 0.
Step 4: In a velocity-time graph, this motion forms a triangle starting from 0, rising to V, and falling back to 0.
Step 5: The average speed for a journey that uniformly accelerates from 0 to a peak velocity V and then uniformly retards to 0 is always exactly half of the peak velocity.
Step 6: Substitute V = aT into the average speed logic: Average speed = V / 2 = aT / 2.
Answer: (c) aT/2
(B) VERY SHORT ANSWER TYPE :
1. Write three equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).
Step 1: The first equation relates final velocity with initial velocity, acceleration, and time.
Step 2: Equation 1: v = u + at.
Step 3: The second equation relates displacement with initial velocity, time, and acceleration.
Step 4: Equation 2: S = ut + (1/2)at².
Step 5: The third equation relates final velocity, initial velocity, acceleration, and displacement.
Step 6: Equation 3: v² = u² + 2aS.
Answer: (i) v = u + at, (ii) S = ut + (1/2)at², (iii) v² = u² + 2aS.
2. Write an expression for the distance S covered in time t by a body which is initially at rest and starts moving with a constant acceleration a.
Step 1: Start with the general equation for distance covered: S = ut + (1/2)at².
Step 2: The phrase "initially at rest" tells us that the initial velocity (u) is zero.
Step 3: Substitute u = 0 into the distance equation.
Step 4: The equation becomes: S = (0 × t) + (1/2)at².
Step 5: Simplify the expression by removing the zero term.
Answer: S = (1/2)at²
(C) LONG ANSWER TYPE :
1. Derive the following equations for a uniformly accelerated motion graphically :
(i) v = u + at
(ii) S = ut + 1/2 at²
(iii) v² = u² + 2aS
where the symbols have their usual meanings.
Step 1: Consider a velocity-time graph for a body with initial velocity 'u' and uniform acceleration 'a'.
Step 2: The graph is an inclined straight line starting from point A (0, u) on the y-axis and reaching point B (t, v) after time 't'.
Step 3: Draw a perpendicular from B to the time axis at point C. Here, OC = t and BC = v.
Step 4: Draw a line AD parallel to the time axis from A meeting BC at D. Here, AD = t, CD = u, and BD = v - u.
Step 5: To derive (i) v = u + at: The slope of the velocity-time line AB represents acceleration (a).
Step 6: Slope a = BD / AD = (v - u) / t.
Step 7: Rearranging this gives v - u = at, which simplifies to v = u + at.
Step 8: To derive (ii) S = ut + 1/2 at²: The distance (S) is the total area under the velocity-time graph (area of trapezium OABC).
Step 9: Area = Area of rectangle OADC + Area of triangle ABD.
Step 10: Area of rectangle OADC = OA × OC = u × t.
Step 11: Area of triangle ABD = 1/2 × AD × BD = 1/2 × t × (v - u). Since (v - u) = at, this becomes 1/2 × t × at = 1/2 at².
Step 12: Adding these areas together gives S = ut + 1/2 at².
Step 13: To derive (iii) v² = u² + 2aS: Calculate the area of the trapezium OABC directly using its formula.
Step 14: Area (S) = 1/2 × (Sum of parallel sides) × height = 1/2 × (OA + BC) × OC.
Step 15: S = 1/2 × (u + v) × t.
Step 16: From the first equation, we know t = (v - u) / a.
Step 17: Substitute this 't' into the area equation: S = 1/2 × (v + u) × [(v - u) / a].
Step 18: This gives S = (v² - u²) / 2a. Rearranging this final equation gives v² = u² + 2aS.
Answer: The derivations are complete using the properties of the velocity-time graph.
(D) NUMERICALS :
1. A body starts from rest with a uniform acceleration of 2 m s⁻². Find the distance covered by the body in 2 s.
Step 1: Identify the given values from the problem.
Step 2: The body starts from rest, so initial velocity (u) = 0 m s⁻¹.
Step 3: The uniform acceleration (a) = 2 m s⁻².
Step 4: The time taken (t) = 2 s.
Step 5: We need to find the distance covered (S).
Step 6: Use the second equation of motion: S = ut + (1/2)at².
Step 7: Substitute the values into the formula: S = (0 × 2) + (1/2) × 2 × (2)².
Step 8: Simplify the first part: 0 × 2 = 0.
Step 9: Simplify the second part: (1/2) × 2 × 4 = 4.
Step 10: Add them together: S = 0 + 4 = 4 m.
Answer: The distance covered by the body is 4 m.
2. A body starts with an initial velocity of 10 m s⁻¹ and acceleration 5 m s⁻². Find the distance covered by it in 5 s.
Step 1: Write down the given information.
Step 2: Initial velocity (u) = 10 m s⁻¹.
Step 3: Acceleration (a) = 5 m s⁻².
Step 4: Time (t) = 5 s.
Step 5: Apply the equation for distance: S = ut + (1/2)at².
Step 6: Put the numbers into the formula: S = (10 × 5) + (1/2) × 5 × (5)².
Step 7: Calculate the first part: 10 × 5 = 50.
Step 8: Calculate the second part: (1/2) × 5 × 25 = 62.5.
Step 9: Add the two parts together: S = 50 + 62.5 = 112.5 m.
Answer: The distance covered is 112.5 m.
3. A vehicle is accelerating on a straight road. Its velocity at any instant is 30 km h⁻¹, after 2 s, it is 33.6 km h⁻¹ and after further 2 s, it is 37.2 km h⁻¹. Find the acceleration of vehicle in m s⁻². Is the acceleration uniform ?
Step 1: Convert the initial velocity to m s⁻¹: 30 km h⁻¹ = 30 × (5/18) = 8.33 m s⁻¹.
Step 2: Convert the velocity after 2s to m s⁻¹: 33.6 km h⁻¹ = 33.6 × (5/18) = 9.33 m s⁻¹.
Step 3: Calculate acceleration for the first 2-second interval: a = (v - u) / t = (9.33 - 8.33) / 2.
Step 4: Calculate the result for first interval: 1.0 / 2 = 0.5 m s⁻².
Step 5: Convert the velocity after another 2s to m s⁻¹: 37.2 km h⁻¹ = 37.2 × (5/18) = 10.33 m s⁻¹.
Step 6: Calculate acceleration for the second 2-second interval: a = (10.33 - 9.33) / 2.
Step 7: Calculate the result for second interval: 1.0 / 2 = 0.5 m s⁻².
Step 8: Compare the accelerations. Since they are both exactly 0.5 m s⁻², the acceleration does not change.
Answer: The acceleration is 0.5 m s⁻². Yes, the acceleration is uniform.
4. A body, initially at rest, starts moving with a constant acceleration 2 m s⁻². Calculate : (i) the velocity acquired and (ii) the distance travelled in 5 s.
Step 1: Identify the given values: u = 0 (at rest), a = 2 m s⁻², t = 5 s.
Step 2: To find the acquired velocity (v), use the first equation of motion: v = u + at.
Step 3: Substitute the values: v = 0 + (2 × 5).
Step 4: Calculate final velocity: v = 10 m s⁻¹.
Step 5: To find the distance travelled (S), use the second equation: S = ut + (1/2)at².
Step 6: Substitute the values: S = (0 × 5) + (1/2) × 2 × (5)².
Step 7: Calculate distance: S = 0 + 1 × 25 = 25 m.
Answer: (i) The velocity acquired is 10 m s⁻¹. (ii) The distance travelled is 25 m.
5. A bullet initially moving with a velocity 20 m s⁻¹ strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target.
Step 1: Note the initial velocity: u = 20 m s⁻¹.
Step 2: Note the final velocity since it comes to rest: v = 0 m s⁻¹.
Step 3: Note the penetration distance and convert it to meters: S = 10 cm = 0.1 m.
Step 4: We need to find acceleration without time, so use the third equation: v² = u² + 2aS.
Step 5: Substitute the values: (0)² = (20)² + 2 × a × 0.1.
Step 6: Simplify the equation: 0 = 400 + 0.2a.
Step 7: Rearrange to solve for a: -0.2a = 400.
Step 8: Calculate acceleration: a = -400 / 0.2 = -2000 m s⁻².
Step 9: Retardation is the negative value of acceleration.
Answer: The retardation caused by the target is 2000 m s⁻².
6. A train moving with a velocity of 20 m s⁻¹ is brought to rest by applying brakes in 5 s. Calculate the retardation.
Step 1: Identify the initial velocity of the train: u = 20 m s⁻¹.
Step 2: Identify the final velocity since it is brought to rest: v = 0 m s⁻¹.
Step 3: Identify the time taken to stop: t = 5 s.
Step 4: Use the acceleration formula: a = (v - u) / t.
Step 5: Substitute the values: a = (0 - 20) / 5.
Step 6: Calculate the acceleration: a = -20 / 5 = -4 m s⁻².
Step 7: Retardation is defined as negative acceleration.
Answer: The retardation is 4 m s⁻².
7. A train travels with a speed of 60 km h⁻¹ from station A to station B and then comes back with a speed 80 km h⁻¹ from station B to station A. Find : (i) the average speed, and (ii) the average velocity of train.
Step 1: Let the distance between station A and station B be 'x' km.
Step 2: Time taken to travel from A to B: t1 = Distance / Speed = x / 60 hours.
Step 3: Time taken to travel back from B to A: t2 = Distance / Speed = x / 80 hours.
Step 4: Total distance travelled for the round trip = x + x = 2x km.
Step 5: Total time taken = t1 + t2 = (x / 60) + (x / 80).
Step 6: Find a common denominator to add the times: (4x + 3x) / 240 = 7x / 240 hours.
Step 7: Calculate average speed = Total Distance / Total Time = 2x / (7x / 240).
Step 8: Simplify the fraction: Average speed = (2 × 240) / 7 = 480 / 7 ≈ 68.57 km h⁻¹.
Step 9: To find average velocity, look at displacement. Since the train returns to station A, displacement is 0.
Step 10: Average velocity = Total Displacement / Total Time = 0 / Total Time = 0.
Answer: (i) Average speed is 68.57 km h⁻¹. (ii) Average velocity is 0.
8. A train is moving with a velocity of 90 km h⁻¹. It is brought to stop by applying the brakes which produce a retardation of 0.5 m s⁻². Find : (i) the velocity after 10 s, and (ii) the time taken by the train to come to rest.
Step 1: Convert the initial velocity to m s⁻¹: 90 km h⁻¹ = 90 × (5/18) = 25 m s⁻¹.
Step 2: Note the acceleration. Since it is retardation, a = -0.5 m s⁻².
Step 3: To find velocity after 10 s, use the first equation: v = u + at.
Step 4: Substitute the values: v = 25 + (-0.5 × 10).
Step 5: Calculate final velocity: v = 25 - 5 = 20 m s⁻¹.
Step 6: To find the time taken to come to rest, set final velocity (v) to 0.
Step 7: Rearrange the first equation for time: t = (v - u) / a.
Step 8: Substitute the values: t = (0 - 25) / -0.5.
Step 9: Calculate the time: t = 50 s.
Answer: (i) The velocity after 10 s is 20 m s⁻¹. (ii) Time taken to come to rest is 50 s.
9. A car travels a distance 100 m with a constant acceleration and average velocity of 20 m s⁻¹. The final velocity acquired by the car is 25 m s⁻¹. Find : (i) the initial velocity and (ii) acceleration of car.
Step 1: The formula for Average Velocity under constant acceleration is (u + v) / 2.
Step 2: We are given Average Velocity = 20 m s⁻¹ and final velocity v = 25 m s⁻¹.
Step 3: Substitute the values into the average velocity formula: 20 = (u + 25) / 2.
Step 4: Multiply both sides by 2 to solve for u: 40 = u + 25.
Step 5: Calculate the initial velocity: u = 40 - 25 = 15 m s⁻¹.
Step 6: We also know Average Velocity = Total Distance / Total Time. Let's find Total Time (t).
Step 7: t = Distance / Average Velocity = 100 / 20 = 5 s.
Step 8: Now use the formula for acceleration: a = (v - u) / t.
Step 9: Substitute the values: a = (25 - 15) / 5.
Step 10: Calculate acceleration: a = 10 / 5 = 2 m s⁻².
Answer: (i) The initial velocity is 15 m s⁻¹. (ii) The acceleration is 2 m s⁻².
10. When brakes are applied to a bus, the retardation produced is 25 cm s⁻² and the bus takes 20 s to stop. Calculate : (i) the initial velocity of bus, and (ii) the distance travelled by bus during this time.
Step 1: Convert retardation into standard S.I. units: 25 cm s⁻² = 0.25 m s⁻².
Step 2: Because it is retardation, the acceleration (a) = -0.25 m s⁻².
Step 3: The bus stops, meaning the final velocity (v) = 0 m s⁻¹.
Step 4: The time taken to stop (t) = 20 s.
Step 5: Use the first equation of motion to find initial velocity: v = u + at.
Step 6: Substitute the values: 0 = u + (-0.25 × 20).
Step 7: Calculate initial velocity: 0 = u - 5, which means u = 5 m s⁻¹.
Step 8: Use the second equation to find distance: S = ut + (1/2)at².
Step 9: Substitute values: S = (5 × 20) + (1/2) × (-0.25) × (20)².
Step 10: Calculate the distance: S = 100 - (0.125 × 400) = 100 - 50 = 50 m.
Answer: (i) The initial velocity is 5 m s⁻¹. (ii) The distance travelled is 50 m.
11. A body moves from rest with a uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.
Step 1: Identify the given conditions: u = 0 (from rest), S = 270 m, t = 3 s.
Step 2: Use the distance equation to find the uniform acceleration: S = ut + (1/2)at².
Step 3: Substitute the known values: 270 = (0 × 3) + (1/2) × a × (3)².
Step 4: Simplify the equation: 270 = 0 + 4.5a.
Step 5: Solve for acceleration: a = 270 / 4.5 = 60 m s⁻².
Step 6: Now, we need to find the velocity at t = 10 s.
Step 7: Use the first equation of motion: v = u + at.
Step 8: Substitute the new time and calculated acceleration: v = 0 + (60 × 10).
Step 9: Calculate the final velocity: v = 600 m s⁻¹.
Answer: The velocity of the body at 10 s is 600 m s⁻¹.
12. A body moving with a constant acceleration travels the distances 3 m and 8 m respectively in 1 s and 2 s. Calculate : (i) the initial velocity, and (ii) the acceleration of body.
Step 1: Write the equation of motion for the first distance: S = ut + (1/2)at².
Step 2: For S = 3 m and t = 1 s, substitute values: 3 = u(1) + (1/2)a(1)².
Step 3: Simplify to form Equation 1: u + 0.5a = 3.
Step 4: Write the equation of motion for the second distance.
Step 5: For S = 8 m and t = 2 s, substitute values: 8 = u(2) + (1/2)a(2)².
Step 6: Simplify the equation: 8 = 2u + 2a. Divide by 2 to get Equation 2: u + a = 4.
Step 7: Subtract Equation 1 from Equation 2 to eliminate 'u': (u + a) - (u + 0.5a) = 4 - 3.
Step 8: Solve for 'a': 0.5a = 1, which means a = 2 m s⁻².
Step 9: Substitute the value of 'a' back into Equation 2: u + 2 = 4.
Step 10: Solve for 'u': u = 2 m s⁻¹.
Answer: (i) The initial velocity is 2 m s⁻¹. (ii) The acceleration is 2 m s⁻².
13. A car travels with a uniform velocity of 25 m s⁻¹ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find : (i) the distance which the car travels before the brakes are applied, (ii) the retardation, and (iii) the distance travelled by the car after applying the brakes.
Step 1: Calculate the distance travelled before applying brakes (uniform velocity).
Step 2: Distance 1 = Velocity × Time = 25 m s⁻¹ × 5 s = 125 m.
Step 3: Analyze the second phase when brakes are applied. Initial velocity for this phase is 25 m s⁻¹.
Step 4: The car comes to rest, so final velocity is 0 m s⁻¹. Time taken is 10 s.
Step 5: Calculate acceleration: a = (v - u) / t = (0 - 25) / 10 = -2.5 m s⁻².
Step 6: Retardation is the negative of acceleration, which is 2.5 m s⁻².
Step 7: Calculate the distance travelled while braking. Use average velocity = (u + v) / 2.
Step 8: Distance 2 = Average Velocity × Time = [(25 + 0) / 2] × 10.
Step 9: Calculate Distance 2: 12.5 × 10 = 125 m.
Answer: (i) Distance before brakes is 125 m. (ii) Retardation is 2.5 m s⁻². (iii) Distance after brakes is 125 m.
14. A space craft flying in a straight course with a velocity of 75 km s⁻¹ fires its rocket motors for 6.0 s. At the end of this time, its speed is 120 km s⁻¹ in the same direction. Find : (i) the space craft's average acceleration while the motors were firing, (ii) the distance travelled by the space craft in the first 10 s after the rocket motors were started, the motors having been in action for only 6.0 s.
Step 1: Note that velocities are in km s⁻¹. Initial velocity (u) = 75 km s⁻¹, final velocity (v) = 120 km s⁻¹, time (t) = 6 s.
Step 2: Calculate average acceleration: a = (v - u) / t.
Step 3: Substitute the values: a = (120 - 75) / 6 = 45 / 6 = 7.5 km s⁻².
Step 4: Calculate distance travelled during the 6 seconds of firing using S = ut + (1/2)at².
Step 5: Substitute values: S1 = (75 × 6) + (1/2) × 7.5 × (6)².
Step 6: Calculate S1 = 450 + 135 = 585 km.
Step 7: For the remaining 4 seconds (to make up the first 10 s), the motors are off. The craft travels at a constant 120 km s⁻¹.
Step 8: Calculate distance for this coasting phase: S2 = Velocity × Time = 120 × 4 = 480 km.
Step 9: Calculate total distance travelled in 10 s: Total S = S1 + S2 = 585 + 480 = 1065 km.
Answer: (i) The average acceleration is 7.5 km s⁻². (ii) The distance travelled is 1065 km.
15. A train starts from rest and accelerates uniformly at a rate of 2 m s⁻² for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find : (i) the maximum velocity reached, (ii) the retardation in the last 50 s, (iii) the total distance travelled, and (iv) the average velocity of the train.
Step 1: Analyze Phase 1 (Acceleration). Starts from rest (u=0), a = 2 m s⁻², t = 10 s.
Step 2: Find maximum velocity reached at the end of Phase 1: v_max = u + at = 0 + (2 × 10) = 20 m s⁻¹.
Step 3: Analyze Phase 3 (Braking). Initial velocity is 20 m s⁻¹, final velocity is 0, time = 50 s.
Step 4: Calculate retardation for Phase 3: a = (v - u) / t = (0 - 20) / 50 = -0.4 m s⁻². So retardation is 0.4 m s⁻².
Step 5: Calculate distance for Phase 1: S1 = (1/2)at² = (1/2) × 2 × (10)² = 100 m.
Step 6: Calculate distance for Phase 2 (Constant speed): S2 = Velocity × Time = 20 × 200 = 4000 m.
Step 7: Calculate distance for Phase 3 (Braking): S3 = Average Velocity × Time = [(20 + 0) / 2] × 50 = 500 m.
Step 8: Calculate total distance: Total S = S1 + S2 + S3 = 100 + 4000 + 500 = 4600 m.
Step 9: Calculate total time: Total t = 10 + 200 + 50 = 260 s.
Step 10: Calculate average velocity for the entire trip: Total Distance / Total Time = 4600 / 260 ≈ 17.69 m s⁻¹.
Answer: (i) Maximum velocity is 20 m s⁻¹. (ii) Retardation is 0.4 m s⁻². (iii) Total distance is 4600 m. (iv) Average velocity is 17.69 m s⁻¹.
Competency Focussed Questions
(E) ASSERTION - REASON TYPE QUESTIONS :
(Choose the answer based on the codes given below.)
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are True and R is not the correct explanation of A
(c) Assertion is false but reason is true
(d) Assertion is true but reason is false
(i) Assertion (A) : The displacement of a body can be zero even if the distance travelled by it is not zero.
Reason (R) : Displacement is the shortest distance from the initial to the final position of body.
Step 1: Evaluate the Assertion (A). If a person walks around a block and returns to the start, distance is non-zero, but displacement is zero. Thus, A is true.
Step 2: Evaluate the Reason (R). The definition of displacement is indeed the shortest distance between initial and final positions. Thus, R is true.
Step 3: Determine if R explains A. Because displacement only measures the gap between start and end, it perfectly explains why returning to the start results in zero displacement. R correctly explains A.
Answer: (a) Both A and R are true and R is the correct explanation of A
(ii) Assertion (A) : For a given time interval, average velocity and average speed can have different values.
Reason (R) : Speed is a scalar quantity whereas velocity is a vector quantity.
Step 1: Evaluate the Assertion (A). Average speed depends on total path length (distance), while average velocity depends on straight line change (displacement). They can easily be different. A is true.
Step 2: Evaluate the Reason (R). It correctly states that speed is a scalar (magnitude only) and velocity is a vector (magnitude and direction). R is true.
Step 3: Determine if R explains A. The difference in their vector/scalar nature is exactly why distance and displacement differ, causing average speed and velocity to differ. R correctly explains A.
Answer: (a) Both A and R are true and R is the correct explanation of A
(iii) Assertion (A) : The slope of displacement-time graph gives the velocity.
Reason (R) : Velocity is the product of displacement and time.
Step 1: Evaluate the Assertion (A). The slope (y/x) of a displacement-time graph is indeed velocity. A is true.
Step 2: Evaluate the Reason (R). Velocity is defined as displacement divided by time (ratio), not their product. R is false.
Step 3: Since A is true and R is false, we look for the corresponding option.
Answer: (d) Assertion is true but reason is false
(iv) Assertion (A) : The velocity-time graph of a body in motion can be a straight line parallel to the time axis.
Reason (R) : This is because the motion of body is with non-uniform velocity.
Step 1: Evaluate the Assertion (A). A velocity-time graph parallel to the time axis means velocity is constant over time. A is true.
Step 2: Evaluate the Reason (R). A constant velocity is known as uniform velocity, not non-uniform velocity. R is false.
Step 3: Since A is true and R is false, we select the corresponding option.
Answer: (d) Assertion is true but reason is false
(v) Assertion (A) : The acceleration-time graph of a body is a straight line parallel to time axis.
Reason (R) : This depicts the motion of freely falling body under gravity.
Step 1: Evaluate the Assertion (A) in the context of the reason. A straight line parallel to the time axis on an acceleration-time graph indicates constant (uniform) acceleration.
Step 2: Evaluate the Reason (R). A freely falling body under gravity experiences a constant acceleration of roughly 9.8 m s⁻². Therefore, it would indeed generate such a graph. R is true.
Step 3: Determine if R explains A. The reason gives a perfectly valid real-world example (free fall) that correctly explains why you would see this specific graph shape. Both are true and R explains A.
Answer: (a) Both A and R are true and R is the correct explanation of A
(vi) Assertion (A) : Acceleration of a moving body is always positive.
Reason (R) : Acceleration is the rate of change of velocity with time.
Step 1: Evaluate the Assertion (A). A moving body can slow down, which results in negative acceleration (retardation). Therefore, acceleration is not always positive. A is false.
Step 2: Evaluate the Reason (R). The strict definition of acceleration is the rate of change of velocity with time. R is true.
Step 3: Since A is false but R is true, we select the corresponding option.
Answer: (c) Assertion is false but reason is true
(F) CASE STUDY BASED QUESTION :
1. A lift starts from rest at the ground floor and moves vertically upward along a straight shaft. During the first 4 s, the lift moves with uniform acceleration and attains a velocity of 2 m s⁻¹. It then moves with uniform velocity for the next 6 s. Finally, it is brought to rest uniformly in 2 s at the top floor.
Assuming the motion to be one-dimensional, answer the following:
(a) Calculate the acceleration of the lift during the first 4 s.
Step 1: Note the values for the first 4 seconds. Initial velocity (u) = 0, Final velocity (v) = 2 m s⁻¹, Time (t) = 4 s.
Step 2: Use the formula: a = (v - u) / t.
Step 3: Substitute the values: a = (2 - 0) / 4.
Step 4: Calculate the result: a = 0.5 m s⁻².
Answer: 0.5 m s⁻²
(b) Calculate the total height between the ground floor and the top floor.
Step 1: The total height is equal to the total distance covered in all three phases of motion.
Step 2: Phase 1 (Acceleration): Distance S1 = Average velocity × time = [(0 + 2) / 2] × 4 = 1 × 4 = 4 m.
Step 3: Phase 2 (Uniform Velocity): Distance S2 = Velocity × time = 2 × 6 = 12 m.
Step 4: Phase 3 (Retardation): Distance S3 = Average velocity × time = [(2 + 0) / 2] × 2 = 1 × 2 = 2 m.
Step 5: Total height = S1 + S2 + S3 = 4 + 12 + 2 = 18 m.
Answer: 18 m
(c) Draw the velocity-time graph for the complete motion and mark the regions of acceleration, retardation and zero acceleration.
Step 1: On a graph with Velocity on the Y-axis and Time on the X-axis, plot the starting point at (0,0).
Step 2: Draw a straight diagonal line upwards from (0,0) to (4, 2). Mark this region as "Acceleration".
Step 3: Draw a horizontal straight line from (4, 2) to (10, 2). Mark this region as "Zero Acceleration" (or uniform velocity).
Step 4: Draw a straight diagonal line downwards from (10, 2) to (12, 0). Mark this region as "Retardation".
Answer: (The graph consists of a trapezium with an upward slope for 4s, flat roof for 6s, and downward slope for 2s).
(d) Use the graph to calculate the average velocity of the lift during the upward journey.
Step 1: The average velocity is the total displacement divided by the total time.
Step 2: From part (b), we know the total displacement (height) is 18 m.
Step 3: The total time for the journey is 4 s + 6 s + 2 s = 12 s.
Step 4: Calculate Average Velocity = 18 / 12.
Step 5: Simplify the result: 1.5 m s⁻¹.
Answer: 1.5 m s⁻¹
(e) If the lift now moves downward following the same pattern, how would the signs of acceleration and displacement change?
Step 1: When moving downward, the direction of motion reverses compared to the upward journey.
Step 2: By convention, if upward is positive, downward becomes negative.
Step 3: As the lift starts moving downward and speeds up, its velocity is becoming more negative, meaning acceleration is negative.
Step 4: The overall change in position is downward, returning to the ground, making the displacement negative.
Answer: Both acceleration (during the initial speed-up phase) and overall displacement would take negative signs.
(f) “The lift has maximum acceleration when it starts moving.” Is this statement correct? Give reason.
Step 1: Let's calculate the magnitude of acceleration for the starting phase.
Step 2: Starting acceleration (a) = 0.5 m s⁻² (calculated in part a).
Step 3: Let's calculate the magnitude of acceleration (retardation) for the stopping phase.
Step 4: Stopping acceleration = (0 - 2) / 2 = -1 m s⁻². The magnitude here is 1 m s⁻².
Step 5: Compare the magnitudes: 1 m s⁻² (when stopping) is greater than 0.5 m s⁻² (when starting).
Step 6: Therefore, the statement is incorrect. The lift experiences maximum acceleration (in terms of magnitude) when it is coming to a stop.
Answer: No, the statement is incorrect because the retardation when stopping (1 m s⁻²) is greater in magnitude than the acceleration when starting (0.5 m s⁻²).
Quick Navigation:
Quick Review Flashcards - Click to flip and test your knowledge!
Question
What is the definition of a physical quantity?
Answer
A quantity that can be measured.
Question
Which two parameters are needed to completely express a scalar quantity?
Answer
The unit of measurement and the numerical value of the quantity.
Question
Term: Scalar quantity
Answer
Definition: A physical quantity expressed only by its magnitude. Example: Mass (5.0\,kg).
Question
List three parameters required to express a vector quantity completely.
Answer
Numerical value, unit, and direction.
Question
How are vector quantities symbolically represented in writing?
Answer
By an English letter with an arrow above it (e.g., \vec{v}) or by a bold English letter (e.g., \mathbf{v}).
Question
In vector notation, what does a negative sign typically imply?
Answer
The reverse or opposite direction.
Question
Give three examples of scalar quantities mentioned in the text.
Answer
Mass, length, and time.
Question
Give three examples of vector quantities mentioned in the text.
Answer
Displacement, velocity, and acceleration.
Question
Under what condition is a body said to be at rest?
Answer
If it does not change its position with respect to its immediate surroundings.
Question
Why is 'rest' considered a relative term rather than an absolute one?
Answer
An object may be at rest relative to one observer but in motion relative to another (e.g., a tree relative to a person on a platform vs a moving train).
Question
When can a moving body be assumed to be a 'point particle'?
Answer
When the distance travelled is much larger than the size of the body.
Question
What defines one-dimensional motion (rectilinear motion)?
Answer
Motion of a body along a straight line path.
Question
Term: Distance
Answer
Definition: The total length of the path along which a body moves. Example: The actual path length taken from point A to point B.
Question
What is the S.I. unit of distance and displacement?
Answer
Metre (m).
Question
What is the C.G.S. unit of distance and displacement?
Answer
Centimetre (cm).
Question
Term: Displacement
Answer
Definition: The shortest distance from the initial position to the final position of a body in a specific direction.
Question
How does the magnitude of displacement relate to the distance travelled?
Answer
The magnitude of displacement is either equal to or less than the distance.
Question
Under what specific condition is the magnitude of displacement equal to the distance?
Answer
When the motion is in a fixed direction along a straight line.
Question
Can displacement be zero if the distance travelled is non-zero?
Answer
Yes, if the body returns to its starting point after travelling.
Question
The rate of change of distance with time is defined as _____.
Answer
Speed.
Question
What is the formula for calculating speed v?
Answer
v = \frac{S}{t}, where S is distance and t is time.
Question
What is the S.I. unit of speed?
Answer
Metre per second (m\,s^{-1}).
Question
Define uniform speed.
Answer
When a body covers equal distances in equal intervals of time throughout its motion.
Question
What type of speed is measured by the speedometer of a vehicle?
Answer
Instantaneous speed.
Question
State the formula for average speed.
Answer
\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}.
Question
Term: Velocity
Answer
Definition: The distance travelled per second by a body in a specified direction.
Question
How does velocity differ from speed in terms of quantity type?
Answer
Speed is a scalar quantity, while velocity is a vector quantity.
Question
Under what two conditions are two bodies said to be moving with the same velocity?
Answer
If they move with the same speed and in the same direction.
Question
Define uniform velocity.
Answer
When a body travels equal distances in a particular direction in equal intervals of time.
Question
What are the two ways the velocity of a body can change (variable velocity)?
Answer
By a change in its magnitude (speed), its direction, or both.
Question
Why is the motion of a body in a circular path at constant speed considered variable velocity?
Answer
Because the direction of motion changes continuously at every point.
Question
State the formula for average velocity.
Answer
\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time taken}}.
Question
Is it possible for the average velocity of a body to be zero even if its average speed is not?
Answer
Yes, if the body returns to its initial position, displacement is zero.
Question
The rate of change of velocity with time is defined as _____.
Answer
Acceleration.
Question
What is the S.I. unit of acceleration?
Answer
Metre per second square (m\,s^{-2}).
Question
State the mathematical formula for acceleration a using initial velocity u and final velocity v.
Answer
a = \frac{v - u}{t}.
Question
What is retardation (or deceleration)?
Answer
Negative acceleration, occurring when velocity decreases with time.
Question
What does a positive or negative sign for acceleration indicate in straight-line motion?
Answer
Whether the velocity is increasing (positive) or decreasing (negative) with time.
Question
Define uniform acceleration.
Answer
When equal changes in velocity take place in equal intervals of time.
Question
What is the acceleration produced in a body due to Earth's gravitational attraction called?
Answer
Acceleration due to gravity (g).
Question
What is the average value of g on the Earth's surface?
Answer
9.8\,m\,s^{-2} (or approximately 10\,m\,s^{-2}).
Question
How does the sign of g change based on the direction of vertical motion?
Answer
It is +g when falling downwards (velocity increases) and -g when projected upwards (velocity decreases).
Question
Does the value of g depend on the mass of the falling body?
Answer
No, it is independent of the mass of the body.
Question
What physical quantity is represented by the slope of a displacement-time graph?
Answer
Velocity.
Question
In a displacement-time graph, what does a straight line parallel to the time axis represent?
Answer
The body is stationary (at rest).
Question
If a displacement-time graph is a straight line inclined to the time axis, what does this indicate?
Answer
The body is moving with uniform velocity.
Question
What does a negative slope in a displacement-time graph represent?
Answer
The body is returning towards the starting (or reference) point.
Question
Why can a displacement-time graph never be a straight line parallel to the displacement axis?
Answer
It would mean displacement increases infinitely without any increase in time (infinite velocity).
Question
In a displacement-time graph, what does a curve represent?
Answer
Motion with non-uniform (varying) velocity.
Question
What physical quantity is represented by the slope of a velocity-time graph?
Answer
Acceleration.
Question
What is represented by the area enclosed between a velocity-time graph and the time axis?
Answer
Displacement.
Question
In a velocity-time graph, what does a straight line parallel to the time axis represent?
Answer
Motion with uniform velocity (zero acceleration).
Question
What does a straight line inclined to the time axis with a positive slope on a velocity-time graph represent?
Answer
Uniform acceleration.
Question
What does a straight line inclined to the time axis with a negative slope on a velocity-time graph represent?
Answer
Uniform retardation.
Question
How is the total distance travelled calculated from a velocity-time graph?
Answer
By adding the areas under the curve numerically without regard to the sign of displacement.
Question
What information is obtained from the area enclosed between an acceleration-time graph and the time axis?
Answer
The change in speed (or velocity) of the body.
Question
For a body falling freely from rest, displacement S is directly proportional to the _____ of time t.
Answer
Square (S \propto t^2).
Question
State the first equation of uniformly accelerated motion.
Answer
v = u + at.
Question
State the second equation of uniformly accelerated motion.
Answer
S = ut + \frac{1}{2}at^2.
Question
State the third equation of uniformly accelerated motion.
Answer
v^2 = u^2 + 2aS.