Question 1

A truck driver driving at the speed of  $54\mathrm{km}\mathrm{h}^{-1}$  notices a road sign with a speed limit of  $40\mathrm{km}\mathrm{h}^{-1}$  (Fig. 4.29) for trucks. He slows down to  $36\mathrm{km}\mathrm{h}^{-1}$  in  $36\mathrm{s}$ . What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Accepted Answer

Given:
Initial speed, u = 54 km/h = 54 × (1000/3600) = 15 m/s
Final speed, v = 36 km/h = 36 × (1000/3600) = 10 m/s
Time, t = 36 s

Acceleration, a = (v - u)/t = (10 - 15)/36 = -5/36 m/s² (negative sign indicates deceleration)

Distance travelled, s = ut + (1/2) a t²
= 15 × 36 + 0.5 × (-5/36) × (36)²
= 540 - 0.5 × (5/36) × 1296
= 540 - 0.5 × 180
= 540 - 90 = 450 m

Therefore, the truck travelled 450 meters while slowing down. — Step 1: Convert speeds from km/h to m/s.
Step 2: Calculate acceleration using a = (v - u)/t.
Step 3: Use s = ut + 0.5at² to find distance.
Step 4: Substitute values and simplify to get the distance.

Question 2

A car starts from rest and accelerates uniformly to  $20\mathrm{ms}^{-1}$  in 5 seconds. It then travels at  $20\mathrm{ms}^{-1}$  for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

Accepted Answer

Given:
Initial velocity, u = 0 m/s
Final velocity after acceleration, v = 20 m/s
Time to accelerate, t1 = 5 s
Time at constant velocity, t2 = 10 s
Time to stop, t3 = 6 s

Step 1: Calculate acceleration during first phase:
a = (v - u)/t1 = (20 - 0)/5 = 4 m/s²

Distance during acceleration, s1 = ut1 + 0.5 a t1² = 0 + 0.5 × 4 × 25 = 50 m

Step 2: Distance during constant velocity, s2 = v × t2 = 20 × 10 = 200 m

Step 3: Calculate acceleration during braking:
Initial velocity u = 20 m/s, final velocity v = 0, time t3 = 6 s
Acceleration a = (v - u)/t3 = (0 - 20)/6 = -3.33 m/s²

Distance during braking, s3 = ut3 + 0.5 a t3² = 20 × 6 + 0.5 × (-3.33) × 36 = 120 - 60 = 60 m

Total distance travelled = s1 + s2 + s3 = 50 + 200 + 60 = 310 m — Step 1: Calculate acceleration and distance during acceleration phase.
Step 2: Calculate distance during constant velocity phase.
Step 3: Calculate deceleration and distance during braking phase.
Step 4: Sum all distances to get total distance travelled.

Question 3

A bus is travelling at  $36\mathrm{km}\mathrm{h}^{-1}$  when the driver sees an obstacle  $30\mathrm{m}$  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  $2.5\mathrm{ms}^{-2}$ . Will the bus be able to stop before reaching the obstacle?

Accepted Answer

Given:
Initial speed, u = 36 km/h = 36 × (1000/3600) = 10 m/s
Distance to obstacle, d = 30 m
Reaction time, t_r = 0.5 s
Deceleration, a = -2.5 m/s²

Step 1: Distance travelled during reaction time (no deceleration):
s_r = u × t_r = 10 × 0.5 = 5 m

Remaining distance to stop = 30 - 5 = 25 m

Step 2: Calculate stopping distance under deceleration:
Using v² = u² + 2as, final velocity v = 0 (bus stops)
0 = u² + 2 a s_s
s_s = -u² / (2a) = -(10)² / (2 × -2.5) = 100 / 5 = 20 m

Step 3: Total distance covered before stopping = s_r + s_s = 5 + 20 = 25 m

Since 25 m < 30 m, the bus will be able to stop before reaching the obstacle. — Step 1: Calculate distance travelled during driver's reaction time.
Step 2: Calculate stopping distance using kinematic equation.
Step 3: Add both distances and compare with obstacle distance.
Step 4: Conclude whether bus stops in time.

Question 4

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.

Accepted Answer

An object kept on the Earth cannot be considered to be at rest absolutely because the Earth itself is moving around the Sun. However, relative to the Earth’s surface, the object appears to be at rest as it does not change its position with respect to the Earth. Hence, rest and motion are relative concepts depending on the frame of reference chosen. — The concept of rest or motion depends on the frame of reference. Since Earth moves around the Sun, objects on Earth are also moving in that frame. But relative to Earth, they may be at rest.

Question 5

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.

Accepted Answer

(i) The area under the velocity-time graph during the constant velocity portion represents displacement during that time. This area is a rectangle.

(ii) The area under the graph during decreasing velocity is a trapezium or triangle, representing displacement during deceleration.

To calculate total displacement, sum the areas under the graph for all intervals.

Average acceleration = (final velocity - initial velocity) / total time = (v_f - v_i)/120 s.

Exact numerical values depend on the graph data (Fig. 4.30). — Displacement is the area under the velocity-time graph.
Average acceleration is change in velocity over time.
Shading different areas helps visualize displacement during different motion phases.

Question 6

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 distance she ran based on the graph.

Accepted Answer

Distance travelled = Area under the velocity-time graph.

By estimating the area under the graph in Fig. 4.31 (which may be composed of rectangles, triangles, trapeziums), sum the areas to find total distance.

Exact numerical answer depends on the graph values, but the method is:
1. Divide graph into simple shapes.
2. Calculate area of each shape.
3. Sum all areas to get total distance. — Distance is the integral of velocity over time, represented graphically as area under velocity-time curve.
Estimating area by geometric shapes gives approximate distance.

Question 7

On entering a state highway, a car continues to move with a constant velocity of $6\mathrm{ms}^{-1}$ for 2 minutes and then accelerates with a constant acceleration $1\mathrm{ms}^{-2}$ for 6 seconds. Find the displacement of the car on the state highway in the $2\mathrm{min}6\mathrm{s}$ time interval by drawing a velocity-time graph for its motion.

Accepted Answer

Given:
Constant velocity, v = 6 m/s for t1 = 2 minutes = 120 s
Acceleration, a = 1 m/s² for t2 = 6 s

Displacement during constant velocity, s1 = v × t1 = 6 × 120 = 720 m

Initial velocity for acceleration phase, u = 6 m/s
Final velocity after acceleration, v_f = u + a t2 = 6 + 1 × 6 = 12 m/s

Displacement during acceleration, s2 = ut2 + 0.5 a t2² = 6 × 6 + 0.5 × 1 × 36 = 36 + 18 = 54 m

Total displacement = s1 + s2 = 720 + 54 = 774 m

Velocity-time graph will show a horizontal line at 6 m/s for 120 s, then a straight line increasing from 6 m/s to 12 m/s over 6 s. — Step 1: Calculate displacement during constant velocity phase.
Step 2: Calculate displacement during acceleration phase using kinematic equations.
Step 3: Sum both displacements.
Step 4: Sketch velocity-time graph accordingly.

Question 8

Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of $5\mathrm{ms}^{-1}$ in 5 s. Car B attains a velocity of $3\mathrm{ms}^{-1}$ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

Accepted Answer

Car A:
Initial velocity u = 0, final velocity v = 5 m/s, time t = 5 s
Acceleration a = (v - u)/t = 5/5 = 1 m/s²

Car B:
Initial velocity u = 0, final velocity v = 3 m/s, time t = 10 s
Acceleration a = 3/10 = 0.3 m/s²

Velocities at 5 instants for Car A (every 1 s):
At t=0: 0 m/s
At t=1: 1 m/s
At t=2: 2 m/s
At t=3: 3 m/s
At t=4: 4 m/s
At t=5: 5 m/s

Velocities at 5 instants for Car B (every 2 s):
At t=0: 0 m/s
At t=2: 0.6 m/s
At t=4: 1.2 m/s
At t=6: 1.8 m/s
At t=8: 2.4 m/s
At t=10: 3 m/s

Displacement for Car A in 5 s:
s = ut + 0.5 a t² = 0 + 0.5 × 1 × 25 = 12.5 m

Displacement for Car B in 10 s:
s = 0 + 0.5 × 0.3 × 100 = 15 m

Plot velocity-time graphs with these points and calculate displacement as area under the graph. — Step 1: Calculate acceleration for both cars.
Step 2: Calculate velocities at equal intervals to plot graph.
Step 3: Calculate displacement using s = ut + 0.5at².
Step 4: Plot graphs and verify displacement as area under velocity-time graph.

Question 9

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).

Accepted Answer

Given:
Length of minute hand = radius r = 7 cm = 0.07 m
Time interval = 1.5 hours = 90 minutes

(i) Distance travelled:
The tip of the minute hand moves along the circumference of a circle.
Distance travelled = number of rotations × circumference
Number of rotations = 90 minutes / 60 minutes per rotation = 1.5 rotations
Circumference = 2 \pi r = 2 × 3.1416 × 0.07 = 0.4398 m
Distance = 1.5 × 0.4398 = 0.6597 m

(ii) Displacement:
Displacement is the straight line distance between initial and final position of the tip.
After 1.5 rotations, the tip is at 180° from start (since 1 rotation = 360°, 1.5 rotations = 540°, equivalent to 180°)
Displacement = 2r = 2 × 0.07 = 0.14 m

(iii) Speed = total distance / time
Time = 90 minutes = 5400 s
Speed = 0.6597 / 5400 = 0.000122 m/s

(iv) Velocity = displacement / time
Velocity = 0.14 / 5400 = 0.0000259 m/s

Therefore,
Distance travelled ≈ 0.66 m
Displacement = 0.14 m
Speed ≈ 1.22 × 10^{-4} m/s
Velocity ≈ 2.59 × 10^{-5} m/s — Step 1: Calculate number of rotations in given time.
Step 2: Calculate distance as circumference × rotations.
Step 3: Calculate displacement as chord length for 180° rotation.
Step 4: Calculate speed and velocity by dividing distance and displacement by time respectively.

Question 10

Which of the following best describes linear motion?

Accepted Answer

Motion along a straight line — Linear motion is defined as the motion of an object along a straight line. This is the simplest form of motion, such as a car moving on a straight highway or a ball falling vertically.

Question 11

In describing the position of an object moving in a straight line, why do we need to specify a reference point?

Accepted Answer

A reference point is needed to describe the position of an object because position is defined as the distance and direction of the object from that fixed point. Without a reference point, the position cannot be determined. For example, the starting point of a runner can be taken as the reference point to describe her position at different times. — Position is relative and depends on a fixed reference point. By specifying a reference point, we can measure how far and in which direction the object is located at any instant.

Question 12

Refer to Figure 4.3 which shows a straight line with origin O as reference point. Positions to the right of O are positive and to the left are negative. If an athlete moves from O to point A at +100 m and then back to point B at +40 m, what is the displacement of the athlete from O to B?

Accepted Answer

+40 m — Displacement is the change in position from the initial point to the final point. Since point B is at +40 m from origin O, the displacement from O to B is +40 m.

Question 13

What is the difference between distance and displacement?

Accepted Answer

Distance is the total length of the path traveled by an object and is a scalar quantity having only magnitude. Displacement is the shortest straight-line distance between the initial and final positions of the object with direction and is a vector quantity. For example, if a person walks 3 m east and then 4 m west, the distance is 7 m but the displacement is 1 m east. — Distance measures the entire path length irrespective of direction, while displacement considers only the net change in position with direction.

Question 14

If an athlete runs 100 m forward and then 60 m backward along the same straight line, what is the total distance travelled and the magnitude of displacement?

Accepted Answer

Distance = 160 m, Displacement = 40 m — Total distance is the sum of all distances covered: 100 m + 60 m = 160 m. Displacement is the net change from start to final position: 100 m - 60 m = 40 m.

Question 15

In the vertical motion of a ball thrown upwards from point O, moving up to B and falling back to O, what is the displacement when the ball reaches B if the distance OB is 40 cm?

Accepted Answer

40 cm upward / 40 cm in upward direction — Displacement is the straight-line distance from the initial position to the current position with direction. When the ball reaches B, it is 40 cm above O, so displacement is 40 cm upward.

Describing Motion

Describing Motion — Study Notes

Introduction

Distance and Displacement

Speed

Practice Questions — Describing Motion

All 13 Chapters in Exploration