Class 09 : Motion : NCERT In Text Solution

 Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance.

Suppose a ball starts moving from point A and it returns back at same point A, then the distance will be equal to 20 meters while displacement will be zero.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

Given, side of the square field = 10m

Therefore, perimeter = 10 m x 4 = 40 m

Farmer moves along the boundary in 40s.

Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?

Since in 40 s farmer moves 40 m

∴ in 1s distance covered by farmer $=\frac{40}{40}m=1m$

∴ in 140s distance covered by farmer $=1\times 140m=140m$

Now, number of rotation to cover 140m along the boundry$=\frac{\text{Total distance}}{\text{Perimeter}}$

$\frac{140m}{40m}=3.5$ round

Thus, after 3.5 round farmer will at point C of the field.

∴ Displacement $AC=\sqrt{{\left(10m\right)}^2+{\left(10m\right)}^2}$

$=\sqrt{100m^2+100m^2}$

$=\sqrt{200m^2}$

$=\sqrt{2\times 100m^2}$

$=10\sqrt{2}m$

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Question 3: Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer: None

Question 4: Distinguish between speed and velocity.

Answer: Speed has only magnitude while velocity has both magnitude and direction.

Question 5: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: When distance is equal to displacement.

Question 6: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 7: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion the path of an object will look like a straight line.

Question 8: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 ms-1.

Answer: Here, we have, speed $=3\times {10}^{8}m{s}^{-1}$

Time = 5 minute

$=5\times 60s=300$ second

We know that, Distance = Speed x Time

⇒ Distance $=300\times 3\times {10}^{8}m{s}^{-1}$

$=900\times {10}^{8}m$

$=9\times {10}^{10}m$

Question 9: When will you say a body is in

(i) uniform acceleration?

Answer: When rate of change of motion is same in equal intervals of time.

(ii) non-uniform acceleration?

Answer: When rate of change of motion is not same in equal intervals of time.

Question 10: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Answer: Here we have, u=80 km/h $=80\times \frac{5}{18}=\frac{200}{9}$ m/s

v=60 km/h $=60\times \frac{5}{18}=\frac{50}{3}$ m/s and $t=5s$

∴ Acceleration (a) =?

We know that $v=u+at$

Or, $a=\frac{v-u}{t}$

$=\frac{\frac{50}{3}-\frac{200}{9}}{5}=-\frac{50}{9\times 5}$

$=-\frac{10}{9}$ m s-2

Question 11: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Answer: Here we have,
Initial velocity, u = 0,
Final velocity, v = 40km/h $=40\times \frac{5}{18}=\frac{100}{9}$ m/s
Time (t) = 10 minute = 60 × 10 = 600s
Acceleration (a) =?

We know that $v=u+at$

Or, $a=\frac{v-u}{t}$

$=\frac{100}{9\times 600}=\frac{1}{54}$ m s-2

Question 12: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer:

(a) The slope of the distance-time graph for an object in uniform motion is straight line.

(b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 13: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: When the slope of distance-time graph is a straight line parallel to time axis, the object is moving with uniform motion.

Question 14: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.

Question 15: What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity time graph.

Question 16: A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.

Answer: Here we have, Initial velocity (u) = 0
Acceleration (a) = 0.1ms-2
Time (t) = 2 minute = 120 second

(a) The speed acquired:

We know that, v = u + at

$\Rightarrow v=0+0.1m/s^2\times 120s$

$\Rightarrow v=120m/s$

Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

We know that, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=0\times 120s+\frac{1}{2}\times 0.1m/s^2\times {\left(120s\right)}^2$

$=\frac{1}{2}\times 1440m=720m$

Thus, bus will travel a distance of 720 m in the given time of 2 minute.

Question 17: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.

Answer: Here,we have,

Initial velocity, $u=90km/h$

$=\frac{90\times 1000m}{60\times 60s}=25m/s$

Final velocity $v=0$

Acceleration, $a=-0.5m/s^2$

Thus, distance travelled =?

We know that, $v^2=u^2+2as$

$\Rightarrow 0={(25m/s)}^2+2\times -0.5m/s^2\times s$

$\Rightarrow 0=625m^2{s}^{-2}-1m{s}^{-2}s$

$\Rightarrow 1m{s}^{-2}s=625m^2{s}^{-2}$

$\Rightarrow s=\frac{625m^2{s}^{-2}}{1m{s}^{-2}}=625m$

Therefore, train will go 625 m before it brought to rest.

Question 18: A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?

Answer: Here we have,
Initial velocity, u = 0
Acceleration (a) = 2cm/s2 = 0.02m/s2
Time (t) = 3s
Therefore, Final velocity, v =?

We know that, $v=u+at$

$\therefore v=0+0.02m/s^2\times 3s$

$\Rightarrow v=0.06m/s$

Therefore, the final velocity of trolley will be 0.06m/s after start

Question 19: A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?

Answer: Here we have,
Acceleration, a = 4m/s2
Initial velocity, u =0
Time, t = 10s
Therefore, Distance (s) covered =?

We know that, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=0\times 10s+\frac{1}{2}\times 4m/s^2\times {\left(10s\right)}^2$

$\Rightarrow s=\frac{1}{2}\times 4m/s^2\times 100s^2$

$\Rightarrow s=2\times 100m=200m$

Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.

Question 20: A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer: Here we have,

Initial velocity (u) = 5m/s

Final velocity (v) =0 (Since from where stone starts falling its velocity will become zero)

Acceleration (a) = -10m/s2

(Since given acceleration is in downward direction, i.e. the velocity of the stone is decreasing, thus acceleration is taken as negative)

Height, i.e. Distance, s =?

Time (t) taken to reach the height =?

We know that, $v^2=u^2+2as$

$\Rightarrow 0={(5m/s)}^2+2\times -10m/s^2\times s$

$\Rightarrow 0=25m^2{s}^2-20m/s^2\times s$

$\Rightarrow 20m/s^2\times s=25m^2{s}^2$

$\Rightarrow s=\frac{25m^2{s}^2}{20m/s^2}$

$\Rightarrow s=1.25m$

Now, we know that, $v=u+at$

$\Rightarrow 0=5m{s}^{-1}+(-10m{s}^{-2})\times t$

$\Rightarrow 0=5m{s}^{-1}-10m{s}^{-2}\times t$

$\Rightarrow 10m{s}^{-2}\times t=5m{s}^{-1}$

$\Rightarrow t=\frac{5m{s}^{-1}}{10m{s}^{-2}}$

$\Rightarrow t=\frac{1}{2}s=0.5s$

Thus, stone will attain a height of 1.25m. And time taken to attain this height is 0.5s

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