Page 48 – Think It Over
Q1: How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
Ans: We should maintain a safe distance from the truck ahead so that there is enough time to stop the vehicle if the truck suddenly applies brakes, thus avoiding a collision.
Q2: Does this distance depend upon the speed with which we are moving?
Ans: Yes, this distance depends on the speed of the vehicle. As speed increases, the stopping distance also increases, so a larger distance must be maintained.
Page 51 – Pause and Ponder
Q1. 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?
Ans. The displacement of the athlete will be zero when the athlete returns to the same position from which they started, i.e., back to the origin O. This is because displacement is the net change in position between the starting and stopping points. If the athlete starts at O and returns to O, the net change in position is zero, so displacement = zero.
In that case, the total distance travelled will NOT be zero – it will be the sum of all the paths covered by the athlete. For example, in Fig. 4.4, if the athlete goes from O to A (100 m) and then comes back all the way to O, the total distance = 100 m + 100 m = 200 m, but displacement = 0 m. So displacement is zero but total distance travelled is 200 m (or more, depending on how far the athlete went).
Q2. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement
Ans. Fuel used up in a vehicle depends on (i) Total distance travelled.
Fuel used depends on the total distance travelled, not on displacement. This is because fuel is consumed throughout the entire path of motion, regardless of direction. Displacement only gives the shortest distance between initial and final positions, so it does not represent the actual fuel consumption. For example, if a car goes from home to a market 2 km away and comes back home, its displacement is zero but it has consumed fuel to travel 4 km total.
Q3. 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?
Ans. Yes, the motion of the ball is along a straight line (along the inclined track). Its motion from O to D can be represented using a horizontal line by taking positions along the track as distances from the origin. The values of total distance travelled and magnitude of displacement from O are equal at positions A, B, C, and D because the motion is along a straight line in one direction.
Page 53 – Pause and Ponder
Q4. 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.
Ans.
Given:
- First part: 200 km North in 3 hours
- Second part: 200 km South in 2 hours
Step 1: Calculate total distance travelled
Total distance = 200 km + 200 km = 400 km
Step 2: Calculate total time
Total time = 3 hours + 2 hours = 5 hours
Step 3: Calculate average speed
Average speed = total distance / total time = 400 km / 5 h = 80 km h⁻¹
Step 4: Calculate displacement
You travel 200 km North, then 200 km South – you return to your starting point. Net displacement = 200 km North – 200 km South = 0 km
Step 5: Calculate average velocity
Average velocity = displacement / total time = 0 km / 5 h = 0 km h⁻¹
Q5. 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?
Ans.
(i) The magnitude of average velocity is equal to average speed when:
The object moves in one direction only, without turning back, i.e., when the object moves in a straight line in one direction throughout the entire time interval. In this case, the total distance travelled = the magnitude of displacement, so average speed = magnitude of average velocity.
For example, a car moving from point A to point B along a straight road without reversing – both average speed and magnitude of average velocity are the same.
(ii) The magnitude of average velocity is zero while average speed is not zero when:
The object starts and ends at the same position (returns to the starting point) but has actually travelled some distance. In this case, displacement = 0, so average velocity = 0. But since the object has covered some distance, average speed is not zero. Examples:An athlete running one complete lap on a circular track and returning to the starting line
Revise, Reflect, Refine
Q1. 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?
Ans.
Let Home = H and Shop = S. Distance HS = 250 m.
Journey:
- Home → Shop: 250 m
- Shop → Home: 250 m (came back to get the cloth bag)
- Home → Shop: 250 m (went again)
- Shop → Home: 250 m (came back home)
Total distance travelled = 250 + 250 + 250 + 250 = 1000 m
Displacement from home = 0 m
This is because displacement is the net change in position between the starting and ending points. Since he started and ended at home, the displacement is zero.
Q2. 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.
Ans.
Setting up the problem:
- Ground floor = 0 m (starting point)
- Each floor height = 3 m
- 1st floor = 3 m, 2nd floor = 6 m, 3rd floor = 9 m, 4th floor = 12 m
Journey:
- Ground floor → 4th floor: travels upward = 12 m
- 4th floor → 2nd floor: travels downward = 12 – 6 = 6 m
Part (i): Total vertical distance travelled
Total distance = distance going up + distance coming down = 12 m + 6 m = 18 m
Part (ii): Displacement from starting point
Starting point = Ground floor = 0 m Ending point = 2nd floor = 6 m
Displacement = final position – initial position = 6 m – 0 m = 6 m (upward direction)
The displacement is 6 m in the upward direction, because the student started at the ground floor and ended at the 2nd floor, which is 6 m above the starting point. The direction is upward.
Q3. 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?
Ans.Yes, it is possible. Even if the speed is constant, the scooter can still be accelerating if there is a change in direction of motion. Acceleration depends on change in velocity, and velocity includes both speed and direction. So, when the scooter takes a turn or moves along a curved path, its direction changes, causing acceleration even though the speed remains constant. Example: A scooter moving at constant speed on a circular road is accelerating due to continuous change in direction.
Q4. 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.
Ans.
Given:
- Initial velocity, u = 0 m s⁻¹ (starts from rest)
- Final velocity, v = 24 m s⁻¹
- Time, t = 6 s
Finding average acceleration:
a = (v – u) / t = (24 – 0) / 6 = 24 / 6 = 4 m s⁻²
Finding distance travelled:
Using s = ut + ½at²:
s = (0)(6) + ½ × 4 × (6)²
s = 0 + ½ × 4 × 36
s = 2 × 36 = 72 m
Verification using v² = u² + 2as:
24² = 0² + 2 × 4 × s
576 = 8s
s = 576 / 8 = 72 m
Answers:
- Average acceleration = 4 m s⁻²
- Distance travelled in 6 s = 72 m
Q5. 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.
Ans.
Given:
- Initial velocity, u = 28 m s⁻¹
- Final velocity, v = 0 m s⁻¹ (stops)
- Distance, s = 98 m
Finding acceleration:
Using v² = u² + 2as:
0² = 28² + 2 × a × 98
0 = 784 + 196a
196a = -784
a = -784 / 196 = -4 m s⁻²
The negative sign indicates the acceleration is opposite to the direction of motion (it is deceleration/retardation). The magnitude of acceleration = 4 m s⁻².
Finding time taken:
Using v = u + at:
0 = 28 + (-4) × t
4t = 28
t = 28 / 4 = 7 s
Answers:
- Acceleration of the motorbike = -4 m s⁻² (deceleration/retardation of 4 m s⁻²)
- Time taken to stop = 7 s
Q6. 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.
Ans.No, objects A and B never have equal velocity. The position-time graph of both objects is a straight line, which indicates constant velocity. Velocity is given by the slope of the graph. Since the slope of line A is steeper than that of line B, object A has a greater velocity than object B. As both slopes remain constant and different throughout, their velocities never become equal at any instant.
Q7. 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.
Ans.
(i) Since both A and B start at the same position and end at the same position after 10 s, their displacement is the same. Average velocity = displacement / time, and since displacement and time are the same for both, their average velocities are equal. This statement is TRUE.
(ii) Average speed = total distance travelled / time. The total distance covered depends on the actual path length, not just the start and end positions. A (curved graph) has travelled more total distance during its motion than B (it moved faster earlier and covered more ground). So their average speeds are NOT necessarily equal. This statement is FALSE.
(iii) Object A (curved line) covers more total path length than B during the 10 s interval – it moves fast in the first half and slower later, but overall covers more distance. Object B covers less total distance. Therefore, the average speed of A is greater than (not lower than) that of B. This statement is FALSE.
(iv) The average speed of A over the 10 s interval is greater than that of B, because A covers more total distance than B over the same 10 s. This statement is TRUE.
Correct answers: (i) and (iv)
Q8. 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 (Fig. 4.29). 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.
Ans.
Given:
- Initial velocity, u = 54 km h⁻¹ = 54 × (1000/3600) = 54/3.6 = 15 m s⁻¹
- Final velocity, v = 36 km h⁻¹ = 36/3.6 = 10 m s⁻¹
- Time, t = 36 s
- Acceleration is constant
Finding acceleration:
a = (v – u) / t = (10 – 15) / 36 = -5 / 36 m s⁻²
Finding distance travelled using s = ut + ½at²:
s = 15 × 36 + ½ × (-5/36) × (36)²
s = 540 + ½ × (-5/36) × 1296
s = 540 + (-5/36) × 648
s = 540 – 90
s = 450 m
Verification using v² = u² + 2as:
10² = 15² + 2 × (-5/36) × s
100 = 225 – (10/36) × s
(10/36) × s = 125
s = 125 × 36 / 10 = 450 m
Answer: Distance travelled = 450 m
Q9. 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.
Ans.
The motion has three phases:
Phase 1: Acceleration from rest to 20 m s⁻¹ in 5 s
- u = 0 m s⁻¹, v = 20 m s⁻¹, t = 5 s
Distance s₁ = ½ (u + v) × t = ½ × (0 + 20) × 5 = ½ × 20 × 5 = 50 m
(Or using s = ut + ½at²: a = 20/5 = 4 m s⁻², s = 0 + ½ × 4 × 25 = 50 m ✓)
Phase 2: Constant velocity at 20 m s⁻¹ for 10 s
- u = v = 20 m s⁻¹, t = 10 s
Distance s₂ = v × t = 20 × 10 = 200 m
Phase 3: Braking from 20 m s⁻¹ to rest in 6 s
- u = 20 m s⁻¹, v = 0 m s⁻¹, t = 6 s
Distance s₃ = ½ (u + v) × t = ½ × (20 + 0) × 6 = ½ × 20 × 6 = 60 m
Total distance travelled = s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m
Q10. 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?
Ans.
Given:
- Initial speed of bus, u = 36 km h⁻¹ = 36/3.6 = 10 m s⁻¹
- Reaction time = 0.5 s (during this time, the bus continues at 10 m s⁻¹)
- Deceleration after braking, a = -2.5 m s⁻²
- Obstacle is 30 m ahead
Step 1: Distance covered during reaction time (before brakes are applied)
During reaction time, bus moves at 10 m s⁻¹ for 0.5 s:
d₁ = u × reaction time = 10 × 0.5 = 5 m
Step 2: Distance covered after brakes are applied until the bus stops
After braking: u = 10 m s⁻¹, v = 0, a = -2.5 m s⁻²
Using v² = u² + 2as:
0 = 10² + 2 × (-2.5) × s
0 = 100 – 5s
5s = 100
s = 20 m
d₂ = 20 m
Step 3: Total distance covered before stopping
Total stopping distance = d₁ + d₂ = 5 + 20 = 25 m
Conclusion: The total distance covered by the bus from when the driver sees the obstacle to when the bus completely stops is 25 m. Since the obstacle is 30 m away and the bus stops after only 25 m, yes, the bus will be able to stop before reaching the obstacle. The bus stops 30 – 25 = 5 m before the obstacle.
Q11. 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.
Ans: Rest and motion are relative and depend on the reference point. If the Sun is taken as the reference point, the Earth is moving around it, so an object on Earth is also in motion. However, if the Earth is taken as the reference point, the object does not change its position and is considered at rest. Thus, an object can be at rest with respect to one reference point and in motion with respect to another.
Q12. The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) 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.
Ans.
Shading:
(i) The displacement during constant velocity is represented by the rectangular area under the graph from 20 s to 100 s.
(ii) The displacement when velocity is decreasing is represented by the triangular area from 100 s to 120 s.
Calculating displacement:
= Area of triangle (0-20 s) + Area of rectangle (20-100 s) + Area of trapezium (100-120 s)
= (1/2 × 20 × 3) + (80 × 3) + (1/2 × (3 + 2) × 20)
= 30 + 240 + 50
= 320 m
Calculating average acceleration over 120 s:
Average acceleration = (v – u)/t
= (2 – 0)/120
= 1/60 m/s² ≈ 0.017 m/s²o.
Q13. 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.
Ans.
The running distance = area enclosed under the velocity-time graph and the time axis.
Estimating area:
Phase 1 (0 to 2 h): Approximately a trapezium with parallel sides 5.0 and 7.5 km h⁻¹, width 2 h: Area₁ = ½ × (5.0 + 7.5) × 2 = ½ × 12.5 × 2 = 12.5 km
Phase 2 (2 to 4 h): Rectangle with height 7.5 km h⁻¹, width 2 h: Area₂ = 7.5 × 2 = 15 km
Phase 3 (4 to 6 h): Approximately a trapezium with parallel sides 7.5 and some lower value. If it decreases linearly to about 5 km h⁻¹: Area₃ = ½ × (7.5 + 5.0) × 2 = ½ × 12.5 × 2 = 12.5 km
Estimated total running distance ≈ 12.5 + 15 + 12.5 = 40 km
Q14. 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.
Ans.
Given:
- Phase 1: constant velocity = 6 m s⁻¹ for 2 min = 120 s
- Phase 2: acceleration = 1 m s⁻² for 6 s (starting from 6 m s⁻¹)
Velocity-time graph :
Final velocity at end of Phase 2:Using v = u + at: v = 6 + 1 × 6 = 12 m s⁻¹
Displacement calculation (Area under graph):
Phase 1 displacement (rectangle): s₁ = v × t = 6 × 120 = 720 m
Phase 2 displacement (trapezium): s₂ = ½ × (u + v) × t = ½ × (6 + 12) × 6 = ½ × 18 × 6 = 54 m
(Or using s = ut + ½at²: s₂ = 6 × 6 + ½ × 1 × 36 = 36 + 18 = 54 m )
Total displacement = s₁ + s₂ = 720 + 54 = 774 m
Q15. 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 (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
Ans.
Step 1: Find accelerations
For Car A: a_A = (v – u) / t = (5 – 0) / 5 = 1 m s⁻²
For Car B: a_B = (v – u) / t = (3 – 0) / 10 = 0.3 m s⁻²
Step 2: Calculate velocities at five instants of time for each carTime (s)Velocity of A (m s⁻¹) v = 1×tVelocity of B (m s⁻¹) v = 0.3×t000220.6441.2661.8882.410103.0
Velocity-time graph description:
- Both lines start at origin (0, 0)
- Car A: steeper straight line (reaches 5 m s⁻¹ at 5 s, 10 m s⁻¹ at 10 s)
- Car B: less steep straight line (reaches 3 m s⁻¹ at 10 s)
Step 3: Calculate displacement
For Car A in 5 s (time interval of Car A reaching 5 m s⁻¹): s_A = ½ × (u + v) × t = ½ × (0 + 5) × 5 = 12.5 m
For Car B in 10 s (time interval of Car B reaching 3 m s⁻¹): s_B = ½ × (u + v) × t = ½ × (0 + 3) × 10 = 15 m
(These values equal the areas of the triangles under each line in the velocity-time graph.)
Answers:
- Acceleration of A = 1 m s⁻², Acceleration of B = 0.3 m s⁻²
- Displacement of A in 5 s = 12.5 m
- Displacement of B in 10 s = 15 m
Q16. 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 (Fig. 4.32).
Ans.
Given:
- Length of minute’s hand (radius), R = 7 cm
- Time interval = 6:00 PM to 7:30 PM = 90 minutes = 5400 s
In 60 minutes, the minute’s hand completes 1 full revolution.In 90 minutes, it completes 1.5 revolutions (1½ rounds).
Part (i): Distance travelled
The minute’s hand tip moves along the circumference of a circle with radius 7 cm.
Circumference = 2πR = 2 × (22/7) × 7 = 2 × 22 = 44 cm
In 1 complete revolution, distance = 44 cm. In 1.5 revolutions, distance = 1.5 × 44 = 66 cm
Part (ii): Displacement
At 6:00 PM, the minute’s hand is pointing at 12 (top of the clock). After 90 minutes (1.5 revolutions), the minute’s hand is pointing at 6 (bottom of the clock).
The displacement is the straight-line distance from the tip’s initial position (at 12, i.e., directly above the centre) to its final position (at 6, i.e., directly below the centre).
Initial position of tip: top of the clock, at 7 cm above the centre. Final position of tip: bottom of the clock, at 7 cm below the centre.
Straight-line distance between these two points = 7 + 7 = 14 cm
Direction: from the 12 position downward to the 6 position, i.e., 14 cm directed downward (from top to bottom of clock face).
Part (iii): Speed
Speed = total distance / time = 66 cm / 5400 s = 66 / 5400 m/s
= 0.01222 cm s⁻¹ ≈ 1.22 × 10⁻² cm s⁻¹
Converting: 0.66 m / 5400 s ≈ 1.22 × 10⁻⁴ m s⁻¹
Part (iv): Velocity (Average velocity)
Average velocity = displacement / time = 14 cm / 5400 s
= 0.00259 cm s⁻¹ ≈ 2.59 × 10⁻³ cm s⁻¹
Direction: from the 12 o’clock position toward the 6 o’clock position (straight downward through the centre of the clock face).
Summary of answers:
- Distance travelled = 66 cm
- Displacement = 14 cm (directed from 12 o’clock position toward 6 o’clock position)
- Speed ≈ 1.22 × 10⁻² cm s⁻¹ (or ≈ 1.22 × 10⁻⁴ m s⁻¹)
- Average velocity ≈ 2.59 × 10⁻³ cm s⁻¹ (directed from 12 toward 6 position)