Motion - Questions and Answers

Text Questions

Exercise Questions

Text Questions

Page 74 Questions

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

Yes, an object can have zero displacement even after moving through a distance. This happens when the object returns to its starting point.

Example: A person walks from point A to point B (a distance of 5 km) and then returns back to point A. The total distance travelled is 10 km, but the displacement is zero because the initial and final positions 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?

Side of square field = 10 m

Perimeter of square = 4 × 10 = 40 m

Time to complete one round = 40 s

Total time = 2 minutes 20 seconds = 140 seconds

Number of rounds completed = 140/40 = 3.5 rounds

After 3.5 rounds, the farmer is at the opposite corner of the square

Displacement = diagonal of square = √(10² + 10²) = √(100 + 100) = √200 = 10√2 ≈ 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.

Neither (a) nor (b) is true for displacement.

Explanation:

(a) is false because displacement can be zero when an object returns to its starting point.
(b) is false because the magnitude of displacement is always less than or equal to the distance travelled, but never greater.

Page 76 Questions

1. Distinguish between speed and velocity.

Speed:

Speed is the distance travelled by an object per unit time
It is a scalar quantity (has only magnitude)
It is always positive
SI unit: m/s

Velocity:

Velocity is the displacement of an object per unit time
It is a vector quantity (has both magnitude and direction)
It can be positive, negative or zero
SI unit: m/s

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

The magnitude of average velocity equals average speed when:

The object moves in a straight line
The object moves in a single direction without changing direction
When distance travelled equals 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, it covers equal distances in equal intervals of time. The path can be a straight line (linear motion) or a curved path (circular motion), but the speed remains constant.

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.

Time taken, t = 5 minutes = 5 × 60 = 300 seconds

Speed of light, v = 3 × 10⁸ m/s

Distance = speed × time

Distance = 3 × 10⁸ m/s × 300 s = 9 × 10¹⁰ m

Distance = 9 × 10¹⁰ m = 90,000,000 km

Page 77 Questions

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

(i) A body is in uniform acceleration when its velocity changes by equal amounts in equal intervals of time. The acceleration remains constant.

(ii) A body is in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time. The acceleration varies with time.

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.

Initial velocity, u = 80 km/h = 80 × (1000/3600) = 22.22 m/s

Final velocity, v = 60 km/h = 60 × (1000/3600) = 16.67 m/s

Time, t = 5 s

Acceleration, a = (v - u)/t = (16.67 - 22.22)/5 = -5.55/5 = -1.11 m/s²

The negative sign indicates 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.

Initial velocity, u = 0 (starting from rest)

Final velocity, v = 40 km/h = 40 × (1000/3600) = 11.11 m/s

Time, t = 10 minutes = 10 × 60 = 600 seconds

Acceleration, a = (v - u)/t = (11.11 - 0)/600 = 11.11/600 = 0.0185 m/s²

Page 81 Questions

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

Uniform motion: The distance-time graph is a straight line with constant slope.

Non-uniform motion: The distance-time graph is a curved line with varying slope.

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

When the distance-time graph is a straight line parallel to the time axis, it means the object is at rest. The distance remains constant with time, indicating no motion.

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?

When the speed-time graph is a straight line parallel to the time axis, it means the object is moving with constant speed (uniform motion). The speed does not change with time.

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 of the object.

Page 82 Questions

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.

Initial velocity, u = 0

Acceleration, a = 0.1 m/s²

Time, t = 2 minutes = 120 seconds

(a) Speed acquired, v = u + at = 0 + 0.1 × 120 = 12 m/s

(b) Distance travelled, s = ut + ½at² = 0 + ½ × 0.1 × (120)² = 0.5 × 0.1 × 14400 = 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.

Initial velocity, u = 90 km/h = 90 × (1000/3600) = 25 m/s

Final velocity, v = 0 (brought to rest)

Acceleration, a = -0.5 m/s²

Using v² = u² + 2as

0 = (25)² + 2 × (-0.5) × s

0 = 625 - s

s = 625 m

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?

Initial velocity, u = 0

Acceleration, a = 2 cm/s² = 0.02 m/s²

Time, t = 3 s

Velocity, v = u + at = 0 + 0.02 × 3 = 0.06 m/s

Page 83 Questions

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

Initial velocity, u = 0

Acceleration, a = 4 m/s²

Time, t = 10 s

Distance, s = ut + ½at² = 0 + ½ × 4 × (10)² = 2 × 100 = 200 m

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?

Initial velocity, u = 5 m/s (upward)

Acceleration, a = -10 m/s² (downward)

At maximum height, final velocity v = 0

Using v = u + at

0 = 5 + (-10)t

10t = 5

t = 0.5 s (time to reach maximum height)

Using v² = u² + 2as

0 = (5)² + 2 × (-10) × s

0 = 25 - 20s

20s = 25

s = 1.25 m (maximum height)

Exercise Questions

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?

Diameter of track = 200 m, Radius = 100 m

Time for one round = 40 s

Total time = 2 minutes 20 s = 140 s

Number of rounds completed = 140/40 = 3.5 rounds

Distance covered = 3.5 × circumference = 3.5 × 2πr = 3.5 × 2 × 3.14 × 100 = 2198 m

After 3.5 rounds, the athlete is at the opposite end of the diameter

Displacement = diameter = 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) From A to B:

Distance = 300 m, Time = 2 min 30 s = 150 s

Displacement = 300 m

Average speed = 300/150 = 2 m/s

Average velocity = 300/150 = 2 m/s

(b) From A to C:

Distance = 300 + 100 = 400 m

Time = 150 + 60 = 210 s

Displacement = 300 - 100 = 200 m (from A to C)

Average speed = 400/210 = 1.9 m/s

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 = d km

Time for forward trip = d/20 hours

Time for return trip = d/30 hours

Total distance = 2d km

Total time = d/20 + d/30 = (3d + 2d)/60 = 5d/60 = d/12 hours

Average speed = total distance/total time = 2d/(d/12) = 2d × 12/d = 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?

Initial velocity, u = 0

Acceleration, a = 3.0 m/s²

Time, t = 8.0 s

Distance, s = ut + ½at² = 0 + ½ × 3.0 × (8.0)² = 1.5 × 64 = 96 m

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?

First car: u = 52 km/h = 14.44 m/s, v = 0, t = 5 s

Distance = area under v-t graph = ½ × base × height = ½ × 5 × 14.44 = 36.1 m

Second car: u = 3 km/h = 0.83 m/s, v = 0, t = 10 s

Distance = ½ × 10 × 0.83 = 4.15 m

The first car travelled farther after brakes were applied.

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?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?

(a) B is travelling the fastest (steepest slope)

(b) No, all three are never at the same point on the road

(d) When B passes C, B has travelled approximately 9 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?

Initial velocity, u = 0

Acceleration, a = 10 m/s²

Distance, s = 20 m

Using v² = u² + 2as

v² = 0 + 2 × 10 × 20 = 400

v = 20 m/s (velocity when it strikes the ground)

Using v = u + at

20 = 0 + 10t

t = 2 s (time to strike the ground)

8. The speed-time graph for a car is shown in 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.
(b) Which part of the graph represents uniform motion of the car?

(a) Distance in first 4 seconds = area under graph from 0 to 4 s

Area = ½ × base × height = ½ × 4 × 6 = 12 m

(b) The horizontal part of the graph (from 6 s to 10 s) represents uniform motion where speed is constant at 6 m/s.

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
(b) an object moving with an acceleration but with uniform speed
(c) an object moving in a certain direction with an acceleration in the perpendicular direction

(a) Possible - At the highest point of vertical motion, velocity is zero but acceleration due to gravity is constant (10 m/s² downward)

(b) Possible - Uniform circular motion: speed is constant but acceleration (centripetal) is present due to change in direction

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.

Radius, r = 42250 km

Time period, T = 24 hours = 24 × 3600 = 86400 seconds

Circumference = 2πr = 2 × 3.14 × 42250 = 265,330 km

Speed = distance/time = 265,330 km / 24 h = 11,055.4 km/h

Or in m/s: 265,330,000 m / 86,400 s = 3,070.6 m/s