Question

the speedometer readings of a car are shown below time 9:00 am speedometer readings 36km/h time 9:36 am speedometer readings 72 km/h find the acceleration of car using above information​

Answer 1

Formula used:

a=v−uta=v−ut

2as=v2−u22as=v2−u2,

where aa is constant or average acceleration, vv is final velocity, uu is initial velocity, ss is displacement and tt is time.

Complete step by step answer:

Let us assume that the car is moving along a straight line in a single direction and at a constant acceleration. Then the speed of the car is equal to the velocity of the car. The acceleration of a body is defined as the rate of change in velocity of the body with respect to time. If the body is moving with constant acceleration, then its acceleration is equal to

a=v−uta=v−ut …. (i)

Here, u=36kmh−1=36×103m3600s=10ms−1u=36kmh−1=36×103m3600s=10ms−1.

And v=72kmh−1=72×103m3600s=20ms−1v=72kmh−1=72×103m3600s=20ms−1.

From the table we get that the car is moving with velocity 10ms−110ms−1 at 9:25 am and with velocity 20ms−120ms−1 at 9:45 am. This means that the time taken to increase the velocity from u to v is t=20min=20×60=1200st=20min=20×60=1200s.Substitute these values in (i).

a=20−101200⇒a=101200⇒a=1120ms−2a=20−101200⇒a=101200⇒a=1120ms−2

To find the displacement of the car, we shall use the kinematic equation 2as=v2−u22as=v2−u2 …. (ii).

Substitute the values of a, v and u in (ii).

⇒2(1120)s=(20)2−(10)2⇒2(1120)s=(20)2−(10)2

⇒(160)s=400−100⇒(160)s=400−100

∴s=60×300=18000m=18km∴s=60×300=18000m=18km

Therefore, the acceleration of the car is 1120ms−21120ms−2 and its displacement for that time is 18km.