Think It Over
Question 1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
Answer:
We should maintain a safe distance from the truck ahead. This distance should be enough so that if the truck suddenly applies brakes, our vehicle also gets enough time and space to stop safely.
This safe distance depends on:
- Speed of our vehicle
- Reaction time of the driver
- Condition of brakes
- Road condition, such as dry, wet, or slippery road
- Weight of the vehicle
If we are moving fast, we need more distance to stop. If we are moving slowly, less distance may be enough.
So, we should always keep a proper gap from the vehicle ahead to avoid collision.
Question 2. Does this distance depend upon the speed with which we are moving?
Answer:
Yes, this distance depends on the speed with which we are moving.
When the speed of a vehicle is high, it takes more time and more distance to stop after applying brakes. Therefore, the safe distance from the truck ahead should be larger.
When the speed is low, the vehicle can stop in a shorter distance.
For example, a fast-moving car needs more stopping distance than a slow-moving car. Therefore, higher speed means greater safe distance is needed.
Pause and Ponder — Page 51
Question 1. In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?
Answer:
The displacement of the athlete will be zero when the athlete returns to the starting point.
In Fig. 4.4, the athlete starts from point O, moves to point A, and if she comes back to point O, her final position will be the same as her initial position.
So,
Displacement = Final position − Initial position
Since final position and initial position are the same,
Displacement = 0
If the athlete goes from O to A and returns to O:
Distance from O to A = 100 m
Distance from A to O = 100 m
Total distance travelled = 100 m + 100 m = 200 m
Therefore, the displacement will be zero, but the total distance travelled will be 200 m.
Question 2. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement
Answer:
Fuel used up in a vehicle depends on total distance travelled, not on displacement.
A vehicle uses fuel according to the actual path covered by it. Displacement only tells us the shortest distance between the starting point and the final point. It does not tell us how much path the vehicle actually travelled.
For example, if a vehicle goes from home to a shop and returns home, its displacement is zero because the starting and final positions are the same. But fuel is still used because the vehicle travelled some distance.
Therefore, fuel used depends on:
(i) Total distance travelled
It does not depend on:
(ii) Displacement
Question 3. A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion, a straight line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C and D?
Answer:
Yes, the motion of the ball is a straight line motion because it rolls down a straight inclined track.
Yes, its motion from O to D can be shown using a horizontal line, as in Fig. 4.3. The horizontal line is only a way to represent positions along a straight path.
Given:
OA = 40 cm, AB = 10 cm, BC = 20 cm, CD = 30 cm
So,
At A: distance = displacement = 40 cm
At B: distance = displacement = 40 + 10 = 50 cm
At C: distance = displacement = 40 + 10 + 20 = 70 cm
At D: distance = displacement = 40 + 10 + 20 + 30 = 100 cm
Therefore, the total distance travelled and magnitude of displacement are equal at A, B, C and D, because the ball moves in one direction without turning back.
Pause and Ponder — Page 53
Question 4. During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.
Answer:
Given:
Distance travelled north = 200 km
Time taken = 3 hours
Distance travelled south = 200 km
Time taken = 2 hours
Total distance travelled = 200 km + 200 km = 400 km
Total time taken = 3 h + 2 h = 5 h
Average speed:
Average speed = Total distance travelled ÷ Total time taken
Average speed = 400 km ÷ 5 h = 80 km/h
Now, displacement:
The person drives 200 km north and then 200 km south. So, the final position is the same as the starting position.
Displacement = 0 km
Average velocity:
Average velocity = Displacement ÷ Total time taken
Average velocity = 0 km ÷ 5 h = 0 km/h
Therefore:
Average speed = 80 km/h
Average velocity = 0 km/h
Question 5. Under what condition(s) is the
(i) magnitude of average velocity of an object equal to its average speed?
(ii) magnitude of average velocity of an object zero while its average speed is not zero?
Answer:
(i) Magnitude of average velocity is equal to average speed
The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction.
In this condition:
Total distance travelled = Magnitude of displacement
So,
Average speed = Magnitude of average velocity
Example: A car moving 100 m forward in a straight line without turning back.
(ii) Magnitude of average velocity is zero while average speed is not zero
The magnitude of average velocity is zero when the displacement of the object is zero.
This happens when the object returns to its starting point.
In this case:
Displacement = 0
Average velocity = 0
But the object has travelled some distance, so average speed is not zero.
Example: A person goes from home to a shop and returns home.
Total distance travelled is not zero, but displacement is zero.
Therefore, average velocity is zero while average speed is not zero.
Revise, Reflect, Refine — Chapter 4: Describing Motion Around Us
Question 1. My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Answer:
Distance from home to shop = 250 m
Journey made by father:
Home to shop = 250 m
Shop to home = 250 m
Home to shop again = 250 m
Shop to home again = 250 m
Total distance travelled:
250 + 250 + 250 + 250 = 1000 m
Since father finally comes back home, his final position is the same as his starting position.
So, displacement = 0 m
Therefore:
Total distance travelled = 1000 m
Displacement = 0 m
Question 2. A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find:
(i) the total vertical distance travelled, and
(ii) their displacement from the starting point.
Answer:
Height of each floor = 3 m
The student goes from ground floor to fourth floor.
Vertical distance from ground floor to fourth floor:
4 × 3 = 12 m
Then the student comes down from fourth floor to second floor.
Vertical distance from fourth floor to second floor:
2 × 3 = 6 m
(i) Total vertical distance travelled
Total distance = 12 m + 6 m = 18 m
(ii) Displacement from starting point
Starting point = ground floor
Final point = second floor
Displacement = height of second floor from ground floor
2 × 3 = 6 m upward
Therefore:
Total vertical distance travelled = 18 m
Displacement = 6 m upward
Question 3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
Answer:
Yes, it is possible for the scooter to be accelerating even when its speedometer reading is constant.
The speedometer shows only the speed of the scooter, not the direction of motion. Acceleration occurs when velocity changes. Since velocity includes both speed and direction, a change in direction also means a change in velocity.
For example, if the girl is riding the scooter around a circular path at constant speed, the speedometer reading remains constant, but the direction of velocity keeps changing. Therefore, the scooter is accelerating.
So, a scooter can accelerate even at constant speed if its direction of motion changes.
Question 4. A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Answer:
Given:
Initial velocity, u = 0 m s⁻¹
Final velocity, v = 24 m s⁻¹
Time, t = 6 s
Average acceleration:
a = (v − u) / t
a = (24 − 0) / 6
a = 24 / 6 = 4 m s⁻²
Now, distance travelled:
s = ut + ½at²
s = 0 × 6 + ½ × 4 × 6²
s = 0 + 2 × 36
s = 72 m
Therefore:
Average acceleration = 4 m s⁻²
Distance travelled = 72 m
Question 5. A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
Answer:
Given:
Initial velocity, u = 28 m s⁻¹
Final velocity, v = 0 m s⁻¹
Distance, s = 98 m
Using:
v² = u² + 2as
0² = 28² + 2 × a × 98
0 = 784 + 196a
196a = −784
a = −784 / 196
a = −4 m s⁻²
The negative sign shows that acceleration is opposite to the direction of motion.
Now, using:
v = u + at
0 = 28 + (−4)t
4t = 28
t = 28 / 4
t = 7 s
Therefore:
Acceleration = −4 m s⁻²
Time taken to stop = 7 s
Question 6. Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Answer: No, objects A and B never have equal velocity because their position-time graph lines are straight with different slopes. Equal velocity would require both lines to have the same slope (be parallel), which is not the case here.
𝗔𝗻𝗮𝗹𝘆𝘀𝗶𝘀 𝗼𝗳 𝘁𝗵𝗲 𝗴𝗿𝗮𝗽𝗵: On a position-time graph, velocity is represented by the slope (steepness) of the line. A steeper slope means higher velocity.
In Fig. 4.27, both A and B have straight-line graphs (constant velocities). The lines have different slopes, which means different constant velocities. Two straight lines with different slopes never become parallel to each other, so their slopes are always different.
However, the two lines can intersect at a point. At the intersection point, the two objects are at the same position at the same time — but their velocities (slopes) are different.
Question 7. A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
(i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than that of B since B’s speed is lower than A’s in some segments.
Answer:
The correct options are: (i) and (ii)
Explanation: In Fig. 4.28, both objects A and B start from the same position and reach the same final position in 10 seconds.
A has a straight position-time graph, so it moves with constant velocity.
B has a curved position-time graph, so its velocity changes during motion.
However, the graph is a position-time graph, not a route map. The curved line of B does not mean that B travels on a curved path. It only shows that B’s velocity changes with time.
Option (i): Average velocity depends on displacement and time.
Both A and B have the same initial position and same final position. So, their displacement is the same. Since the time taken is also the same, their average velocities are equal.
Therefore, option (i) is correct.
Option (ii): Since both objects move in a straight line from the same initial position to the same final position, they cover equal distance in the same time.
So, their average speeds are also equal.
Therefore, option (ii) is correct.
Option (iii): Object A does not cover a shorter distance than B. Both cover the same distance from the same starting position to the same final position.
Therefore, option (iii) is incorrect.
Option (iv): Although B’s speed is lower than A’s in some parts, B’s speed is higher in other parts. Overall, both cover the same distance in the same time.
Therefore, option (iv) is incorrect.
Final Answer: Correct options: (i) and (ii)
Question 8. A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
Answer:
Given:
Initial velocity, u = 54 km h⁻¹
Final velocity, v = 36 km h⁻¹
Time, t = 36 s
Convert km h⁻¹ into m s⁻¹:
u = 54 × 5/18 = 15 m s⁻¹
v = 36 × 5/18 = 10 m s⁻¹
For constant acceleration, average velocity:
Average velocity = (u + v) / 2
Average velocity = (15 + 10) / 2 = 25 / 2 = 12.5 m s⁻¹
Distance travelled:
s = average velocity × time
s = 12.5 × 36
s = 450 m
Therefore, the truck travelled 450 m while slowing down.
Question 9. A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
Answer:
The motion has three parts.
First part: Car accelerates from rest to 20 m s⁻¹ in 5 s
Given:
u = 0 m s⁻¹
v = 20 m s⁻¹
t = 5 s
Distance travelled:
s₁ = average velocity × time
s₁ = [(0 + 20) / 2] × 5
s₁ = 10 × 5 = 50 m
Second part: Car moves with constant velocity for 10 s
Velocity = 20 m s⁻¹
Time = 10 s
s₂ = vt
s₂ = 20 × 10 = 200 m
Third part: Car slows down from 20 m s⁻¹ to 0 m s⁻¹ in 6 s
u = 20 m s⁻¹
v = 0 m s⁻¹
t = 6 s
s₃ = average velocity × time
s₃ = [(20 + 0) / 2] × 6
s₃ = 10 × 6 = 60 m
Total distance travelled:
s = s₁ + s₂ + s₃
s = 50 + 200 + 60
s = 310 m
Therefore, the total distance travelled by the car is 310 m.
Question 10. A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?
Answer:
Given:
Initial speed of bus = 36 km h⁻¹
Distance of obstacle = 30 m
Reaction time = 0.5 s
Acceleration after braking = −2.5 m s⁻²
Convert speed into m s⁻¹:
36 km h⁻¹ = 36 × 5/18 = 10 m s⁻¹
Distance travelled during reaction time
During reaction time, the bus continues moving with speed 10 m s⁻¹.
Distance = speed × time
Distance = 10 × 0.5 = 5 m
Remaining distance before obstacle:
30 − 5 = 25 m
Braking distance
Using:
v² = u² + 2as
Here:
v = 0 m s⁻¹
u = 10 m s⁻¹
a = −2.5 m s⁻²
0² = 10² + 2 × (−2.5) × s
0 = 100 − 5s
5s = 100
s = 20 m
Total stopping distance:
Reaction distance + braking distance = 5 + 20 = 25 m
The obstacle is 30 m away, so the bus stops after 25 m.
Therefore, yes, the bus will be able to stop before reaching the obstacle. It will stop 5 m before the obstacle.
Question 11. A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
Answer:
Yes, an object kept on the Earth can be considered to be at rest with respect to the Earth.
Rest and motion depend on the reference point.
If we take Earth as the reference point, then an object kept on the ground does not change its position with respect to Earth. So, it is at rest.
But if we take the Sun as the reference point, the same object is moving because Earth is moving around the Sun and the object is moving along with Earth.
Therefore, an object kept on Earth can be considered:
At rest with respect to Earth
In motion with respect to the Sun
This shows that rest and motion are relative.
Question 12. The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas representing the displacement of the cyclist:
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.
Answer:
From Fig. 4.30, the motion can be divided into three parts:
- From 0 s to 20 s: velocity increases from 0 to 3 m s⁻¹
- From 20 s to 80 s: velocity remains constant at 3 m s⁻¹
- From 80 s to 120 s: velocity decreases from 3 m s⁻¹ to 2 m s⁻¹
Shading
For part (i), shade the rectangular area under the graph from 20 s to 80 s.
For part (ii), shade the area under the sloping part of the graph from 80 s to 120 s.
Displacement
Displacement is equal to the area under the velocity-time graph.
From 0 s to 20 s
Area of triangle:
s₁ = ½ × base × height
s₁ = ½ × 20 × 3
s₁ = 30 m
From 20 s to 80 s
Area of rectangle:
s₂ = length × breadth
s₂ = 60 × 3
s₂ = 180 m
From 80 s to 120 s
Area of trapezium:
s₃ = ½ × sum of parallel sides × height
s₃ = ½ × (3 + 2) × 40
s₃ = ½ × 5 × 40
s₃ = 100 m
Total displacement:
s = s₁ + s₂ + s₃
s = 30 + 180 + 100
s = 310 m
Average acceleration in 120 s
Initial velocity = 0 m s⁻¹
Final velocity = 2 m s⁻¹
Time = 120 s
Average acceleration:
a = (v − u) / t
a = (2 − 0) / 120
a = 1 / 60
a = 0.0167 m s⁻²
Therefore:
Displacement = 310 m
Average acceleration = 0.0167 m s⁻²
Question 13. A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the running distance based on the graph.
Answer:
From Fig. 4.31, the velocity is approximately constant at about 7.5 km h⁻¹ for most of the time and slightly decreases near the end.
The total time shown in the graph is 6 hours.
Distance can be estimated from the area under the velocity-time graph.
Approximate average velocity ≈ 7.0 km h⁻¹
Time = 6 h
Distance = average velocity × time
Distance ≈ 7.0 × 6
Distance ≈ 42 km
Therefore, the estimated running distance is about 42 km.
This is close to the standard marathon distance of about 42 km.
Question 14. On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
Answer:
Given:
Constant velocity = 6 m s⁻¹
Time for constant velocity = 2 min = 120 s
Acceleration = 1 m s⁻²
Time of acceleration = 6 s
Total time = 120 s + 6 s = 126 s
First part: Constant velocity
s₁ = vt
s₁ = 6 × 120
s₁ = 720 m
Second part: Accelerated motion
Initial velocity, u = 6 m s⁻¹
Acceleration, a = 1 m s⁻²
Time, t = 6 s
Final velocity:
v = u + at
v = 6 + 1 × 6
v = 12 m s⁻¹
Displacement during acceleration:
s₂ = average velocity × time
s₂ = [(6 + 12) / 2] × 6
s₂ = 9 × 6
s₂ = 54 m
Total displacement:
s = s₁ + s₂
s = 720 + 54
s = 774 m
Therefore, the displacement of the car is 774 m.
In the velocity-time graph, the first part is a horizontal line at 6 m s⁻¹ from 0 s to 120 s. Then the graph rises uniformly from 6 m s⁻¹ to 12 m s⁻¹ from 120 s to 126 s. The area under this graph gives the displacement.
Question 15. Two cars A and B start moving with a constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals.
Answer:
Both cars start from rest.
So, initial velocity = 0 m s⁻¹
For Car A
Final velocity = 5 m s⁻¹
Time = 5 s
Acceleration of A:
a = (v − u) / t
a = (5 − 0) / 5
a = 1 m s⁻²
Displacement of A is area under velocity-time graph:
s = ½ × base × height
s = ½ × 5 × 5
s = 12.5 m
For Car B
Final velocity = 3 m s⁻¹
Time = 10 s
Acceleration of B:
a = (v − u) / t
a = (3 − 0) / 10
a = 0.3 m s⁻²
Displacement of B is area under velocity-time graph:
s = ½ × base × height
s = ½ × 10 × 3
s = 15 m
Values for plotting graph
| Time | Velocity of Car A | Velocity of Car B |
|---|---|---|
| 0 s | 0 m s⁻¹ | 0 m s⁻¹ |
| 1 s | 1 m s⁻¹ | 0.3 m s⁻¹ |
| 2 s | 2 m s⁻¹ | 0.6 m s⁻¹ |
| 3 s | 3 m s⁻¹ | 0.9 m s⁻¹ |
| 4 s | 4 m s⁻¹ | 1.2 m s⁻¹ |
| 5 s | 5 m s⁻¹ | 1.5 m s⁻¹ |
| 10 s | — | 3 m s⁻¹ |
For Car A, draw a straight line from (0, 0) to (5, 5).
For Car B, draw a straight line from (0, 0) to (10, 3).
Therefore:
Displacement of Car A in 5 s = 12.5 m
Displacement of Car B in 10 s = 15 m
Question 16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minute’s hand is 7 cm.
Answer:
Length of minute hand = radius of circular path = 7 cm
Time interval from 6 PM to 7:30 PM = 1 hour 30 minutes = 90 minutes
The minute hand completes one revolution in 60 minutes.
In 90 minutes, number of revolutions:
90 / 60 = 1.5 revolutions
(i) Distance travelled
Distance travelled in one revolution = circumference of circle
= 2πr
= 2 × 22/7 × 7
= 44 cm
Distance travelled in 1.5 revolutions:
= 1.5 × 44
= 66 cm
(ii) Displacement
At 6 PM, the minute hand is at 12.
At 7:30 PM, the minute hand is at 6.
So, the tip of the minute hand reaches the opposite point of the circle.
Displacement = diameter of circle
= 2r
= 2 × 7
= 14 cm downward
(iii) Speed
Time = 90 minutes = 90 × 60 = 5400 s
Speed = distance / time
Speed = 66 cm / 5400 s
Speed = 0.0122 cm s⁻¹
In m s⁻¹:
66 cm = 0.66 m
Speed = 0.66 / 5400
Speed = 1.22 × 10⁻⁴ m s⁻¹
(iv) Velocity
Average velocity = displacement / time
Magnitude of average velocity = 14 cm / 5400 s
= 0.00259 cm s⁻¹ downward
In m s⁻¹:
14 cm = 0.14 m
Average velocity = 0.14 / 5400
= 2.59 × 10⁻⁵ m s⁻¹ downward
Therefore:
Distance travelled = 66 cm
Displacement = 14 cm downward
Speed = 0.0122 cm s⁻¹
Average velocity = 0.00259 cm s⁻¹ downward