NCERT In Text Solution 3

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

`⇒ v = 0 + 0.1m//s^2 xx 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+1/2at^2`

`=>s=0xx120s+1/2xx0.1\ m//s^2xx(120s)^2`

`= 1/2 xx 1440m = 720 m`

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=90\ km//h`

`=(90xx1000m)/(60xx60s)=25\ m//s`

Final velocity `v=0`

Acceleration, `a = -0.5m//s^2`

Thus, distance travelled =?

We know that, `v^2=u^2+2as`

`=> 0 = (25\ m//s)^2 +2 xx-0.5\ m//s^2 xx s`

`=>0=625\ m^2s^(-2) - 1\ m\ s^(-2)s`

`=>1\ ms^(-2)s = 625\ m^2 s^(-2)`

`=>s=(625\ m^2\ s^(-2))/(1\ m\ 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`

`:. v=0+0.02\ m//s^2 xx 3s`

`=>v=0.06\ m//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+1/2 at^2`

`=>s= 0xx10s+1/2xx4\ m//s^2 xx(10s)^2`

`=>s=1/2xx4\ m//s^2 xx 100s^2`

`=>s = 2xx100m = 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`

`=>0= (5\ m//s)^2+2xx-10\ m//s^2 xxs`

`=>0= 25\ m^2s^2 - 20\ m//s^2 xxs`

`=>20\ m//s^2 xxs = 25\ m^2s^2`

`=>s=(25\ m^2s^2)/(20\ m//s^2)`

`=>s = 1.25\ m`

Now, we know that, `v=u+at`

`=>0= 5\ ms^(-1) +(-10\ ms^(-2))xxt`

`=>0=5\ ms^(-1) - 10\ ms^(-2) xxt`

`=>10\ ms^(-2)xxt = 5\ ms^(-1)`

`=>t=(5\ ms^(-1))/(10\ ms^(-2))`

`=>t=1/2s = 0.5\ s`

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

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