Motion

Given,

Diameter of track (d) = 200 m

Circumference of the track = πd = 200π m

So, distance covered in 40 seconds = 200π m

Distance covered in 1 sec = $\dfrac{200π}{40}$

Distance covered in 2 min 20 sec (i.e., 60 + 60 + 20 = 140 sec) = $\dfrac{200π}{40}$ x 140

= $\dfrac{200 \times 22 \times 140}{7 \times 40}$

= 2200 m

Number of rounds completed in 140 sec = $\dfrac{140}{40}$ = 3.5

∴ the final position (with respect to the initial position) is at the opposite end of the circular track. So, the net displacement will be equal to the diameter of the track = 200 m.

Hence, the net distance covered by the athlete is 2200 m and the total displacement of the athlete is 200 m.

(a) From A to B

Given,

Distance covered from A to B = 300 m

Time taken to travel from A to B = 2 min and 30 sec = 60 + 60 + 30 = 150 sec

Displacement from A to B = 300 m

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

Average velocity = $\dfrac{\text{total displacement}}{\text{total time taken}}$

Substituting we get,

Average speed = $\dfrac{300}{150}$ = 2 ms^-1

Average velocity = $\dfrac{300}{150}$ = 2 ms^-1

Hence, average speed from A to B is 2 ms^-1 and average velocity = 2 ms^-1

(b) From A to C

Distance covered from A to C = 300 m + 100 m = 400 m

Time taken from A to C = 2 min 30 sec + 1 min = 60 + 60 + 30 + 60 = 210 sec

Displacement from A to C = 300 m – 100 m = 200 m

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

Average velocity = $\dfrac{\text{total displacement}}{\text{total time taken}}$

Substituting we get,

Average speed = $\dfrac{400}{210}$ = 1.9 ms^-1

Average velocity = $\dfrac{200}{210}$ = 0.952 ms^-1

Hence, average speed from A to C is 1.9 ms^-1 and average velocity = 0.952 ms^-1

Distance travelled to reach the school = distance travelled to reach home = d

Time taken to reach school = t₁

Time taken to reach home = t₂

Average speed = $\dfrac{\text{total distance travelled}}{\text{total time taken}}$

On going to school :

20 = $\dfrac{\text{d}}{\text{t}_1}$ or

t₁ = $\dfrac{\text{d}}{20}$

On return trip :

30 = $\dfrac{\text{d}}{\text{t}_2}$ or

t₂ = $\dfrac{\text{d}}{30}$

Now, the average speed for the entire trip

= $\dfrac{\text{d+d}}{\text{t}_1 + \text{t}_2}$

= $\dfrac{\text{2d}}{\dfrac{\text{d}}{20} + \dfrac{\text{d}}{30}}$

= $\dfrac{\text{2d}}{\dfrac{\text{3d + 2d}}{60}}$

= $\dfrac{\text{2d} \times {60}}{5\text{d}}$

= 24 kmh^-1

∴ Abdul's average speed for the entire trip is 24 kmh^-1.

Given,

Initial velocity (u) = 0

Acceleration (a) = 3 ms^-2

Time period (t) = 8 s

Distance travelled (S) = ?

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$at²

Substituting we get,

S = 0 + $\dfrac{1}{2}$ x 3 x 8 x 8

= 96 m

∴ The motorboat travels a distance of 96 m

(a) The initial speed of the graph is 52 kmh^-1. After the driver applies the brakes, the speed becomes zero.

The x-axis of the graph represents the time and the y-axis of the graph represents the velocity.

In the graph, point A represents the initial velocity of the car, that is, 52 kmh^-1, and point B represents the time when the velocity becomes zero.

The distance travelled by an object is equal to the product of velocity and time.

Therefore, the area under the velocity-time graph represents the distance travelled by the car during the period.

Distance travelled by the car = Area of triangle AOB

(b) The slope of the velocity-time graph is the same throughout. So, the whole graph represents the uniform motion of the car.

(a) As the slope of line B is the greatest, hence, B is travelling at the fastest speed.

(b) No, as the three lines do not intersect at a single point.

(c) As there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals $\dfrac{4}{7}$ km.

As the initial point of object C is 4 graph units away from the origin, its initial distance from the origin is 4 x $\dfrac{4}{7}$ km = $\dfrac{16}{7}$ km

When B passes A, the distance between the origin and C is 8 km.

∴ total distance travelled by C in this time = 8 – $\dfrac{16}{7}$ km = 5.71 km

(d) Total distance travelled by B when it crosses C = 9 x $\dfrac{4}{7}$ = 5.14 km

Given,

Initial velocity (u) = 0

Distance travelled (s) = 20 m

Acceleration (a) = 10 ms^-2

As per the third law of motion,

v² – u² = 2as

Substituting we get,

v² - 0 = 2 x 10 x 20

⇒ v² = 400

⇒ v = $\sqrt{400}$

⇒ v = 20 ms^-1

Hence, ball strikes the ground with a velocity of 20 ms^-1

As per the first equation of motion,

v = u + at

or

t = $\dfrac{\text{v - u}}{\text{a}}$

Substituting we get,

t = $\dfrac{20-0}{10}$ = 2 seconds

∴ the ball reaches the ground after 2 seconds.

(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.

To find this area, we will use the square counting method.

On time axis (i.e., x axis) :

5 squares = 2 units

∴ 1 square = $\dfrac{2}{5}$ units

On speed axis (i.e., y axis) :

3 squares = 2 units

∴ 1 square = $\dfrac{2}{3}$ units

Area of each square = $\dfrac{2}{5}$ x $\dfrac{2}{3}$ = $\dfrac{4}{15}$ sq units.

We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.

Area of fully filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of more than half filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of half filled square = $\dfrac{1}{2}$ x Area of 1 square = $\dfrac{1}{2} \times \dfrac{4}{15}$ = $\dfrac{2}{15}$ sq units

Area of less than half filled square = 0

No. of fully filled squares = 56 [Marked in graph with 'F']

Area of fully filled squares = 56 x $\dfrac{4}{15}$ = 14.93 sq units

No. of more than half filled squares = 4 [Marked in graph with 'M']

Area of more than half filled squares = 4 x $\dfrac{4}{15}$ = 1.06 sq units

No. of half filled squares = 3 [Marked in graph with 'H']

Area of half filled squares = 3 x $\dfrac{4}{15}$ = 0.8 sq units

No. of less than half filled squares = 4 [Marked in graph with 'L']

Area of less than half filled squares = 4 x 0 = 0 sq units

∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units

∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 m

Hence, the car travels 16.4 m in first 4 seconds

(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.

(a) It is possible;
When an object is thrown up vertically into the air, it has constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) It is possible;
When an object moves on a circular path with uniform speed it is still under acceleration because the velocity changes due to continuous change in the direction of motion.

(c) It is possible;
When an object is accelerating in a circular trajectory, the acceleration is in the perpendicular direction.

Given,

Radius of the orbit (r) = 42250 km

∴ circumference of the orbit

= 2 x $\dfrac{22}{7}$ x r

= 2 x $\dfrac{22}{7}$ x 42250

= 265571.42 km

Time taken to complete one revolution in the orbit = 24 hours

Convert hours into sec

1 hour = 3600 sec

So, 24 hour = 24 x 3600 sec

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

Substituting we get,

Speed = $\dfrac{265571.42}{24 \times 3600}$ = 3.07 kms^-1

∴ Speed of the artificial satellite = 3.07 kms^-1.

Yes, an object can have zero displacement if it returns to its initial position.

Example: When a racer runs on a running track, his initial and final position is the same. Hence, the displacement is zero.

Given,

Time period of travel = 2 minutes 20 seconds

Convert min into sec

1 min = 60 sec

So 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 sec

Side of the given square field = 10 m

Hence, the perimeter of a square = 4 x side = 4 x 10 = 40 m

Time taken by the farmer to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of = $\dfrac{40}{40}$ x 1 = 1 m

Hence, distance covered by the farmer in 140 sec = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m

= $\dfrac{\text{total distance}}{\text{perimeter}}$

= $\dfrac{140}{40}$

= 3.5

If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.

∴ Displacement (from Pythagoras theorem)

$\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \\[1em] = \sqrt{(10)^2 + (10)^2} \\[1em] = \sqrt{100 + 100} \\[1em] = \sqrt{200} \\[1em] = 10\sqrt{2} \\[1em] = 10 \times 1.414 \\[1em] = 14.14 \text{ m}$

Hence, the magnitude of displacement = 14.14 m

(a) False.

Reason — The displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) False.

Reason — The displacement of an object can be equal to, but never greater than the distance travelled.

As average speed is the total distance travelled in a time frame and average velocity is the total displacement in a time frame, so the magnitude of average velocity and average speed will be the equal when the total distance travelled is equal to the displacement.

This happens when an object moves along a straight line in the same direction (without turning back).

An odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

The path of an object in uniform motion is a straight line.

Given,

Time taken (t) = 5 min

Convert min into sec

1 min = 60 sec

So, 5 min = 5 x 60 sec = 300 sec

Speed of the signal = 3 × 10⁸ ms^-1

We know, speed = $\dfrac{\text{distance}}{\text{time}}$

Substituting the values, we get,

3 × 10⁸ = $\dfrac{\text{distance}}{300}$

Distance covered (d) = 3 × 10⁸ x 300 = 9 x 10¹⁰ m.

Hence, the distance of the spaceship from the ground station = 9 x 10¹⁰ m

(i) The acceleration is said to be uniform (or constant) when equal changes in velocity take place in equal intervals of time.

Example — The motion of a body under gravity (e.g., free fall of a body)

(ii) If changes in velocity are not same in the same intervals of time, the acceleration is said to be non-uniform.

Example — The motion of a vehicle on a crowded road.

Given,

Initial velocity (u) = 80 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

80 x $\dfrac{5}{18}$ = 22.22 ms^-1

The final velocity (v) = 60 kmh^-1

60 x $\dfrac{5}{18}$ = 16.66 ms^-1

Time (t) = 5 sec

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{16.66 - 22.22}{5}$ = -1.112 ms^-2

∴ the acceleration of the bus is -1.112 ms^-2.

The negative sign indicates that the velocity of the bus is decreasing.

Given.

Initial velocity (u) = 0

Final velocity (v) = 40 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

40 x $\dfrac{5}{18}$ = 11.11 ms^-1

Time (t) = 10 minute

Convert min into s

1 min = 60 s

10 min = 60 x 10 = 600 s

Acceleration (a) =?

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{11.11-0}{600}$ = 0.0185 ms^-2

∴ the acceleration of the bus is 0.0185 ms^-2.

For uniform motion, the distance-time graph is a straight line.

For non-uniform motion, the distance-time graph is a curve.

When distance-time graph is a straight line parallel to the time axis it means that the object is at the same position as time passes. Hence, the object is at rest.

As there is no change in the speed of the object at any point of time, the object is said to be in uniform motion.

As velocity x time = displacement, the area enclosed between the velocity-time sketch and x-axis (i.e., the time axis) gives the displacement of the body.

(a) Given,

Initial velocity (u) = 0

Acceleration (a) = 0.1 ms^-2

Time = 2 minutes

Convert min into sec

1 min = 60 sec

So, 2 min = 2 x 60 sec = 120 s

v = u + at

Substituting we get,

v = 0 + (0.1 x 120) = 12

Hence, speed acquired = 12 ms^-1

(b) According to the third equation of motion,

2as = v² - u²

or

s = $\dfrac{\text{{v}}^2 - \text{{u}}^2}{2\text{a}}$

Since a = 0.1 ms^-2

v = 12 ms^-1

u = 0

t = 120 s

Substituting we get,

s = $\dfrac{{12}^2 - 0}{2 \times 0.1} = \dfrac{144}{0.2}$ = 720 m.

Hence, the total distance travelled is 720 m

Given,

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1

90 x $\dfrac{5}{18}$= 25 ms^-1

Terminal velocity (v) = 0

Acceleration (a) = -0.5 ms^-2

According to the third equation of motion,

v² – u² = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {25}^2 }{2 \times 0.5} = \dfrac{625}{1}$ = 625 m.

Hence, the train travels 625 m before it is brought to rest

Given,

Initial velocity (u) = 0

Acceleration (a) = 2 cms^-2

Time (t) = 3s

According to the first equation of motion,

v = u + at

Substituting we get,

v = 0 + (2 x 3) = 6 cms^-1

Hence, velocity of the trolley after 3 seconds will be 6 cms^-1

Given,

Initial velocity (u) = 0

Acceleration (a) = 4 ms^-2

Time period (t) = 10 s

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$at²

Substituting we get,

S = 0 + $\dfrac{1}{2}$ x 4 x 10 x 10

= 200 m

∴ The car will cover a distance of 200 m.

Given, initial velocity (u) = 5 ms^-1

Terminal velocity (v) = 0

Acceleration = -10 ms^-2 (as it is in upward direction)

According to the third equation of motion,

v² – u² = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {5}^2 }{2 \times -10} = \dfrac{-25}{-20}$ = 1.25 m.

Hence, distance covered = 1.25 m

According to the first equation of motion,

v = u + at

∴ time taken by the stone to reach a position of rest (maximum height) = $\dfrac{\text{v} - \text{u}}{\text{a}}$

Substituting we get,

$\dfrac{0-5}{-10}$ = 0.5 sec

∴ The stone reaches a maximum height of 1.25 m in 0.5 sec.