MOTION - Q&A

Questions Page 74

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object that has moved through a distance can have zero displacement.
Explanation: Displacement is the shortest distance between the initial and final positions of an object. If an object travels and returns to its starting point, its final position coincides with its initial position.
Example: Consider a person jogging on a circular track. If they start at point A, run one complete circle, and return to point A, the distance covered is equal to the circumference of the track (2πr), but the displacement is zero because the starting and ending points are the same.

2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Given:
Side of square field = 10 m
Perimeter of square = 4 × 10 = 40 m
Time taken to cover the boundary (40 m) = 40 s
Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s

Step 1: Calculate the number of rounds.
Speed = Distance / Time = 40 m / 40 s = 1 m/s
Total distance covered in 140 s = Speed × Time = 1 × 140 = 140 m
Number of rounds = Total distance / Perimeter = 140 / 40 = 3.5 rounds

Step 2: Determine final position.
After 3.5 rounds, the farmer will be at the diagonally opposite corner from the starting point.
If the square is ABCD and he starts at A, after 3 complete rounds he is at A. After another 0.5 round, he reaches point C (opposite to A).

Step 3: Calculate displacement.
Displacement is the length of the diagonal AC.
Using Pythagoras theorem in triangle ABC:
AC² = AB² + BC²
AC² = 10² + 10² = 100 + 100 = 200
AC = √200 = 10√2 m
Using √2 ≈ 1.414, Displacement = 10 × 1.414 = 14.14 m

3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

(a) False. Displacement can be zero if the initial and final positions are the same.
(b) False. Displacement is the shortest path between two points, while distance is the actual path length. The magnitude of displacement can be equal to distance (in straight-line motion) or less than distance, but never greater.

Questions Page 76

1. Distinguish between speed and velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

The magnitude of average velocity is equal to the average speed only when an object moves in a straight line and in the same direction (without turning back). In this case, the total distance travelled is equal to the magnitude of displacement.

3. What does the odometer of an automobile measure?

The odometer of an automobile measures the total distance travelled by the vehicle.

4. What does the path of an object look like when it is in uniform motion?

When an object is in uniform motion, its path is a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m s^-1.

Given:
Time (t) = 5 minutes = 5 × 60 = 300 s
Speed (v) = 3 × 10⁸ m/s

Formula: Distance = Speed × Time

Calculation:
Distance = (3 × 10⁸) × 300
Distance = 900 × 10⁸ m
Distance = 9 × 10¹⁰ m

The distance of the spaceship from the ground station is 9 × 10¹⁰ meters.

Questions Page 77

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

(i) Uniform Acceleration: A body is in uniform acceleration if its velocity changes by equal amounts in equal intervals of time. Example: A free-falling body.
(ii) Non-uniform Acceleration: A body is in non-uniform acceleration if its velocity changes by unequal amounts in equal intervals of time. Example: A car driving in heavy traffic.

2. A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus.

Given:
Initial velocity (u) = 80 km/h = 80 × (5/18) = 22.22 m/s
Final velocity (v) = 60 km/h = 60 × (5/18) = 16.67 m/s
Time (t) = 5 s

Formula: Acceleration (a) = (v - u) / t

Calculation:
a = (16.67 - 22.22) / 5
a = -5.55 / 5
a = -1.11 m s^-2

The acceleration is -1.11 m s^-2 (which is deceleration).

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration.

Given:
Initial velocity (u) = 0 (starting from rest)
Final velocity (v) = 40 km/h = 40 × (5/18) = 11.11 m/s
Time (t) = 10 minutes = 10 × 60 = 600 s

Formula: Acceleration (a) = (v - u) / t

Calculation:
a = (11.11 - 0) / 600
a = 11.11 / 600
a ≈ 0.0185 m s^-2

The acceleration is 0.0185 m s^-2.

Questions Page 81

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

- For Uniform Motion: The distance-time graph is a straight line inclined to the time axis. This shows that distance increases linearly with time.
- For Non-uniform Motion: The distance-time graph is a curved line. This shows that the speed is changing with time.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

If the distance-time graph is a straight line parallel to the time axis, it means that the distance of the object is not changing with time. Therefore, the object is at rest (stationary).

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

If the speed-time graph is a straight line parallel to the time axis, it means the speed remains constant over time. Therefore, the object is undergoing uniform motion (zero acceleration).

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

The area under the velocity-time graph represents the displacement (or distance travelled in a straight line) of the object during that time interval.

Questions Page 82

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m s^-2
Time (t) = 2 minutes = 2 × 60 = 120 s

(a) Find Speed (v):
Using the first equation of motion: v = u + at
v = 0 + (0.1 × 120)
v = 12 m/s

(b) Find Distance (s):
Using the second equation of motion: s = ut + (1/2)at²
s = (0 × 120) + 0.5 × 0.1 × (120)²
s = 0 + 0.05 × 14400
s = 720 m

Answer: (a) Speed = 12 m/s, (b) Distance = 720 m.

2. A train is travelling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of -0.5 m s^-2. Find how far the train will go before it is brought to rest.

Given:
Initial velocity (u) = 90 km/h = 90 × (5/18) = 25 m/s
Final velocity (v) = 0 m/s (brought to rest)
Acceleration (a) = -0.5 m s^-2

Find Distance (s):
Using the third equation of motion: v² - u² = 2as
(0)² - (25)² = 2 × (-0.5) × s
-625 = -1 × s
s = 625 m

The train will travel 625 meters before stopping.

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?

Given:
Initial velocity (u) = 0 (assumed starting from rest)
Acceleration (a) = 2 cm s^-2
Time (t) = 3 s

Find Velocity (v):
Using v = u + at
v = 0 + (2 × 3)
v = 6 cm/s

The velocity will be 6 cm/s.

4. A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?

Given:
Initial velocity (u) = 0 m/s (from start)
Acceleration (a) = 4 m s^-2
Time (t) = 10 s

Find Distance (s):
Using s = ut + (1/2)at²
s = (0 × 10) + 0.5 × 4 × (10)²
s = 0 + 2 × 100
s = 200 m

The distance covered is 200 meters.

5. A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given:
Initial velocity (u) = 5 m/s
Final velocity (v) = 0 m/s (at the highest point)
Acceleration (a) = -10 m s^-2 (acting downwards, against motion)

(a) Find Height (s):
Using v² - u² = 2as
(0)² - (5)² = 2 × (-10) × s
-25 = -20s
s = -25 / -20 = 1.25 m

(b) Find Time (t):
Using v = u + at
0 = 5 + (-10 × t)
10t = 5
t = 5 / 10 = 0.5 s

Answer: Height = 1.25 m, Time = 0.5 s.

Exercises

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given:
Diameter = 200 m, Radius (r) = 100 m
Time for 1 round = 40 s
Total Time = 2 min 20 s = 140 s

Step 1: Calculate Rounds
Number of rounds = Total Time / Time for 1 round = 140 / 40 = 3.5 rounds.

Step 2: Calculate Distance
Circumference of track = 2πr = 2 × (22/7) × 100 ≈ 628.57 m
Total Distance = 3.5 × Circumference
Total Distance = 3.5 × 2πr = 3.5 × 200π = 700π ≈ 2200 m

Step 3: Calculate Displacement
After 3.5 rounds, the athlete is at the diametrically opposite point from the start.
Displacement = Diameter of the track
Displacement = 200 m

Answer: Distance = 2200 m, Displacement = 200 m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?

(a) Motion from A to B:
Distance = 300 m
Time = 2 min 30 s = 150 s
Average Speed = Distance / Time = 300 / 150 = 2 m/s
Average Velocity = Displacement / Time = 300 / 150 = 2 m/s (Straight line)

(b) Motion from A to C (A → B → C):
Total Distance = AB + BC = 300 + 100 = 400 m
Total Time = 150 s + 1 min (60 s) = 210 s
Average Speed = 400 / 210 = 1.90 m/s

Total Displacement = AC = AB - BC = 300 - 100 = 200 m
Average Velocity = 200 / 210 = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h^-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h^-1. What is the average speed for Abdul's trip?

Let distance to school be x km.
Time for trip 1 (t₁) = Distance / Speed = x / 20
Time for trip 2 (t₂) = Distance / Speed = x / 30

Total Distance = x + x = 2x
Total Time = t₁ + t₂ = x/20 + x/30 = (3x + 2x) / 60 = 5x / 60 = x / 12

Average Speed = Total Distance / Total Time
Average Speed = 2x / (x / 12) = 2 × 12 = 24 km/h

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time?

Given:
u = 0 m/s
a = 3.0 m s^-2
t = 8.0 s

Formula: s = ut + (1/2)at²
s = (0 × 8) + 0.5 × 3 × (8)²
s = 0 + 1.5 × 64
s = 96 m

Answer: The boat travels 96 meters.

5. A driver of a car travelling at 52 km h^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h^-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Car A:
u = 52 km/h = 52 × (5/18) = 14.44 m/s
t = 5 s, v = 0
Distance = Area under graph (Triangle) = (1/2) × Base × Height
Distance A = 0.5 × 5 × 14.44 = 36.1 m

Car B:
u = 3 km/h = 3 × (5/18) = 0.83 m/s
t = 10 s, v = 0
Distance B = 0.5 × 10 × 0.83 = 4.15 m

Conclusion: The first car (travelling at 52 km/h) travelled farther.

6. Fig 7.10 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
Object B is fastest because its slope (steepness of the line) is the highest.

(b) Are all three ever at the same point on the road?
No, because all three lines do not intersect at a single common point.

(c) How far has C travelled when B passes A?
(Determined from graph readings - approximate)
When B passes A (intersection of B and A lines), look down to the time axis and then up to line C, then across to distance axis.
Answer is approximately 6 km to 7 km (Value depends on precise graph reading, often cited as ~6.5 km or calculated based on grid squares).

(d) How far has B travelled by the time it passes C?
(Determined from intersection of B and C)
Read the Y-axis value at the intersection point.
Answer is approximately 5 km to 6 km (Often cited as ~5.7 km).

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?

Given:
u = 0 m/s
s = 20 m
a = 10 m s^-2

(a) Find Final Velocity (v):
v² - u² = 2as
v² - 0 = 2 × 10 × 20
v² = 400
v = 20 m/s

(b) Find Time (t):
v = u + at
20 = 0 + 10t
t = 2 s

8. The speed-time graph for a car is shown is Fig. 7.11.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
Distance = Area under the curve up to t=4.
The shape is roughly a quarter circle or can be estimated by counting squares.
(Approximation): Area ≈ (1/2) × Base × Height doesn't apply perfectly to curves, but if estimating as a triangle: 0.5 × 4 × 6 = 12 m. By counting squares accurately, it is often around 12 m.

(b) Which part of the graph represents uniform motion of the car?
The flat, horizontal part of the graph (from t=6s to t=10s) represents uniform motion where speed is constant.

9. State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity
Possible. Example: A ball thrown vertically upwards reaches its highest point where its velocity is zero, but it still has a constant acceleration due to gravity (g = 9.8 m s^-2) acting downwards.

(b) an object moving with an acceleration but with uniform speed
Possible. Example: An object moving in a circular path with uniform speed. It has acceleration (centripetal) because the direction of velocity changes continuously.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction
Possible. Example: An airplane moving horizontally while gravity acts vertically downwards (perpendicularly). Or a car turning a curve (velocity is tangential, acceleration is radial/perpendicular).

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Given:
Radius (r) = 42250 km
Time (t) = 24 hours

Formula: Speed (v) = Distance / Time = 2πr / t

Calculation:
v = (2 × 3.14 × 42250) / 24
v = 265330 / 24
v ≈ 11055.4 km/h

To convert to km/s:
v = 11055.4 / 3600 ≈ 3.07 km/s