# Motion : NCERT Intext Questions Page 100
Q.1     An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Sol.     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.

Q.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?
Sol. 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 mTherefore, in 1s distance covered by farmer = ${{40} \over {40}}m = 1m$
Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m
Now, number of rotation to cover 140 along the boundry  =  ${{Total\,dis\tan ce} \over {Perimeter}}$
$= {{140m} \over {40m}} = 3.5\,\,round$
Thus after 3.5 round farmer will at point C of the field.
Therefore, Displacment $AC = \sqrt {{{\left( {10m} \right)}^2} + {{\left( {10m} \right)}^2}}$
$= \sqrt {100{m^2} + 100{m^2}}$
$= \sqrt {200{m^2}}$
$= \sqrt {2 \times 100{m^2}}$
$= 10\sqrt 2 m$ Thus, after 2 minute 20 second the displacement of farmer will be equal to $10\sqrt 2 m$ north east from initial position.

Q.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.
Sol.     None

Page 102

Q.1     Distinguish between speed and velocity.
Sol.     Speed has only magnitude while velocity has both magnitude and direction.

Q.2     Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Sol.     The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity.

Q.3     What does the odometer of an automobile measure?
Sol.     In automobiles, odometer is used to measure the distance.

Q.4     What does the path of an object look like when it is in uniform motion?
Sol.     In the case of uniform motion the path of an object will look like a straight line.

Q.5     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, $3 \times {10^8}m{s^{ - 1}}$.
Sol.     Here we have, speed = $3 \times {10^8}m{s^{ - 1}}$
Time = 5 minute = $5 \times 60s = 300$ second
We know that, Distance = Speed × Time
$\Rightarrow$ Distance = $3 \times {10^8}m{s^{ - 1}} \times 300s = 1800 \times {10^8}m = 1.8 \times {10^{11}}m$

Page 103

Q.1     When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Sol.     (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.
(ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Q.2     A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Sol.     Here we have, u = 80 km/h, v = 60km/h, t = 5s
Therefore, acceleration, a = ?
We know that, v = u + at
$\Rightarrow$ 60 km/ h = 80km/h + a × 5s
$\Rightarrow$ 60km/h – 80km/h = a × 5s
$\Rightarrow$ – 20km/h = a × 5s
$\Rightarrow$ a = $- {{20km/h} \over {5s}}$
$\Rightarrow$ a = – 4km/h/s
Therefore, Acceleration = –4 km/h/s or, –1.1 $m/{s^2}$

Q. 3     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.
Sol.      Here we have,
Initial velocity, u = 0,
Final velocity, v = 40km/h = 11.11m/s
Time (t) = 10 minute = 60 x 10=600s
Acceleration (a) =?
We know that, v = u + at
$\Rightarrow$ 40km/h = 0km/h + a × 10m
$\Rightarrow$ 11.11 m/s = a × 600s
$\Rightarrow$ a = ${{11.11m/s} \over {600s}} = 0.0185m/{s^2}$

Page 107

Q.1     What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Sol.      (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.

Q.2     What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Sol.      When the slope of distance-time graph is a straight line parallel to time axis, the object is moving with uniform motion.

Q.3    What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Sol.      When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.

Q.4    What is the quantity which is measured by the area occupied below the velocity-time graph?
Sol.     The quantity of distance is measured by the area occupied below the velocity time graph.

Page 109

Q.1     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.
Sol.     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
⇒ v = 0 + 0.1m/s2 x 120 s
⇒ v = 120 m/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, $s = ut + {1 \over 2}a{t^2}$
$= 0 \times 120s + {1 \over 2} \times 0.1m/{s^2} \times {\left( {120s} \right)^2}$
$= {1 \over 2} \times 14400m = 7200m\,\,or\,\,7.2km$
Thus, bus will tavel a distance of 7200 m or 7.2 km in the given time of 2 minute.

Page 110

Q.2     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.
Sol.     Here,we have,
Initial velocity, u = 90km/h
$= {{90 \times 1000m} \over {60 \times 60s}} = 25m/s$
Final velocity, v = 0
Acceleration,a = – 0.5 $m/{s^2}$
Therefore, distance travelled = ?
We know tha, ${v^2} = {u^2} + 2as$
$\Rightarrow 0 = {\left( {25m/s} \right)^2} + 2 \times - 0.5m/{s^2} \times s$
$\Rightarrow 0 = 625{m^2}{s^{ - 2}} - 1m{s^{ - 2}}s$
$\Rightarrow 1m{s^{ - 2}}s = 625{m^2}{s^{ - 2}}$ $s = {{625{m^2}{s^{- 2}}} \over {1m{s^{ - 2}}}} = 625m$
Therefore, train will go 625 m be fore it brought to rest.

Q.3     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?
Sol.     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 + 0.02${m/{s^2}}$× 3s
$\Rightarrow$ v = 0.06 m/s
Therefore the final velocity of trolley will be 0.06 m/s after start.

Q.4     A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Sol.     Here we have,
Acceleration, a = 4m/s2
Initial velocity, u = 0
Time, t = 10s
Therefore, Distance (s) covered =?
We know that,  $s = ut + {1 \over 2}a{t^2}$
$\Rightarrow s = 0 \times 10s + {1 \over 2} \times 4m/{s^2} \times {\left( {10s} \right)^2}$
$\Rightarrow s = {1 \over 2} \times 4m/{s^2} \times 100{s^2}$
$\Rightarrow s = 2 \times 100m = 200m$
Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.

Q. 5     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?
Sol.      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, ${v^2} = {u^2} + 2as$
$\Rightarrow 0 = {\left( {5m/s} \right)^2} + 2 \times - 10m/{s^2} \times s$
$\Rightarrow 0 = 25{m^2}{s^2} - 20m/{s^2} \times s$
$\Rightarrow 20m/{s^2} \times s = 25{m^2}{s^2}$ $s = {{25{m^2}{s^2}} \over {20m/{s^2}}}$
$\Rightarrow s = 1.25m$
Now, we know that , v = u + at
$\Rightarrow 0 = 5m{s^{ - 1}} - \left( {10m{s^{ - 2}}} \right) \times t$
$\Rightarrow 0 = 5m{s^{ - 1}} - 10m{s^{ - 2}} \times t$
$\Rightarrow 10m{s^{ - 2}} \times t = 5m{s^{ - 1}}$
$\Rightarrow t = {{5m{s^{ - 1}}} \over {10m{s^{ - 2}}}} = {1 \over 2}s = 0.5s$ Thus, stone will attain a height of 1.25m And time taken to attain the height is 0.5s

• nishat

sorry but regarding

Q.5 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, 3×108ms−1.
Sol. Here we have, speed = 3×108ms−1
Time = 5 minute = 5×60s=300 second
We know that, Distance = Speed × Time
⇒ Distance = 3×108ms−1×300s=1800×108m=1.8×1011m

but 3*300=900 or the above please clarify.

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