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?
Step 1: Distance from home to shop
Distance = 250 m
Step 2: Shop to home
Distance = 250 m
Step 3: Home to shop again
Distance = 250 m
Step 4: Shop to home again
Distance = 250 m
Total Distance Travelled
250+250+250+250 = 1000m
Therefore,
Total distance travelled = 1000 m
Displacement
Displacement is the shortest distance between initial and final position.
Initial position = Home
Final position = Home
So,
Therefore,
Displacement from home = 0 m
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 her 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
Given:
Height of each floor = 3 m
Step 1: Distance travelled upward
Ground floor to fourth floor:
4 x 3 = 12 m
So, upward distance = 12 m
Step 2: Distance travelled downward
Fourth floor to second floor:
Difference in floors:
4 – 2 = 2 floors
Distance downward:
2 x 3 = 6 m
Step 3: Total vertical distance travelled
12 + 6 = 18 m
Therefore,
Total vertical distance travelled = 18 m
Step 4: Displacement from starting point
Starting point = Ground floor
Final position = Second floor
Height of second floor:
2 x 3 = 6 m
Therefore,
Displacement = 6 m upward
3. A girl is riding a scooter and finds that its speedometer reading is 60 km h⁻¹. What can she infer from this reading?
A speedometer measures the instantaneous speed of a moving vehicle.
The reading of 60 km h⁻¹ means that:
- The scooter is moving with a speed of 60 kilometres per hour.
- If the scooter continues at the same speed, it will cover 60 km in 1 hour.
Therefore,
The scooter is moving at a speed of 60 km h⁻¹ at any particlar instant of time.
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.
5. A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration and the time taken to come to a stop.
Given:
Initial velocity, u = 28 m s⁻¹
Final velocity, v = 0 m s⁻¹
Distance, s = 98 m
6. Fig. 4.27 shows the position-time graphs of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B have equal velocity? Justify your answer.
Velocity is given by the slope of the position-time graph.
- In Fig. 4.27, the slope for object A is steeper than the slope for object B.
- A steeper graph means greater velocity.
Therefore:
- Object A has greater velocity than object B.
- Their velocities are not equal.
Therefore,
No, objects A and B do not have equal velocity because their position-time graphs have different slopes.
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).
Options:
(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 velocity 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.
Options:
(i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
Average velocity is:
Average velocity = Displacement/Time
Since:
- Both have same displacement.
- Both take same time (10 s).
Therefore, their average velocities are equal.
Hence,
✔ Option (i) is Correct.
(ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.
Explanation:
This statement is correct because:
Average speed depends on distance.
Both objects A & B covers equal distance during motion.
Hence,
Option (ii) is correct.
(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.
Explanation:
Average speed is = Total distance/ Time
Objects A & B travels a same distance .
Hence,
Option (iii) is Incorrect.
(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.
Explanation:
Even if B’s speed is higher in some parts, overall both A & B covers equal distances in the same time.
So:
Average speed of A & B is equal .
Hence,
Option (iv) is Incorrect.
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 (Fig. 4.29) 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.
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.
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. After 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 hitting the obstacle?
11. A student said, “The Earth moves around the Sun.” In this context, discuss whether an object kept on th Earth can be considered to be at rest?
Rest and motion are relative concepts. An object may appear at rest with respect to one reference point but moving with respect to another.
- An object kept on the Earth appears to be at rest with respect to the Earth and its surroundings.
- However, the Earth itself is continuously moving around the Sun and also rotating on its axis.
- Therefore, the object is actually moving along with the Earth.
For example:
- A book lying on a table seems at rest for a person standing nearby.
- But relative to the Sun, the book is moving because the Earth is moving.
Therefore,
An object kept on the Earth can be considered at rest only with respect to the Earth, but not with respect to the Sun or outer space.
12. The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colors) representing the displacement of the cyclist
(i) While the 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.
(i) Velocity increasing
From graph:
Velocity is increasing from 0 s to 20 s.
(ii) Constant velocity
From graph:
Velocity is constant from 20 s to 100 s.
(iii) Velocity decreasing
From graph:
Velocity decreases from 100 s to 120 s.
Step 1: Displacement
Area under velocity-time graph gives displacement.
13. A girl is preparing for her first marathon by running on a straight path. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts velocity versus time. Estimate the running distance based on the graph.
Step 1: Understand the graph
The graph is a velocity-time graph.
Distance travelled = Area under the velocity-time graph
From the graph: we can analyse that there are different velocities .
i) From A to B , Velocity is 7 km/h
ii) From B to D , Velocity is different at B = 7 km/h and D = 7.5 km/h
iii) From D to F , Velocity is 7.5 km/h
iv) From F to H , Velocity is different at F = 7.5 km/h and H = 7 km/h
v) From H to J , Velocity is different at H = 7 km/h and J = 6.5 km/h
vi) From J to L , Velocity is 6.5 km/h
we have divided the graph into various portions based on these different velocities .
i) Portion OABC (Rectangle )
Area under the curve = 7 × .6 = 4.2 kms
ii) Portion CBDE (Trapezium)
Area under the curve = 1/2 × ( 7 + 7.5 ) × 1 = 7.25 kms
iii) Portion EDFG (Rectangle)
Area under the curve = 1.4 × 7.5 = 10.5 kms
iv) Portion GFHI (Trapezium)
Area under the curve = 1/2 × ( 7 + 7.5 ) × 1.6 = 11.6 kms
v) Portion IHJK (Trapezium)
Area under the curve = 1/2 × ( 7 + 6.5 ) × 1 = 6.75 kms
vi) Portion KJLM (Rectangle)
Area under the curve = 1 × 6.5 = 6.5 kms
Total area under the Velocity Time Graph is 46.8 kms
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 of 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s interval by drawing a velocity-time graph for its motion.
Step 1: Write the given data
First part of motion
- Constant velocity:
v=6 m s−1
Time:
2 minutes = 2 s
Second part of motion
Initial velocity:
u = 6 m s-1
Acceleration:
A = 1 m s−2
Time:
t = 6 s
Step 2: Displacement during first motion
Using formula:
S1 = vt
S1 = 6 × 120
S1 = 720 m
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 graphs, calculate the displacement mentioned in two time intervals (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
Step 1: Given Data
Car A
Initial velocity:
U = 0 m s−1
Final velocity:
v = 5 m s−1
Time:
t = 5 s
Car B
Initial velocity:
u = 0 m s−1
Final velocity:
v = 3 m s−1
Time:
t = 10 s
16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute hand of the wall clock. During the given interval, mention:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minutes hand is 7 cm (fig. 4.32)
