Question

a car is accelerated uniformly from 36 km per hour to 90 km in 3 seconds calculate the acceleration of card the distance covered by car in that time

Answer 1

u=36km/h
=10m/s
v=90km/h
=25m/s
t=3sec

a=(v-u)/t
=15/3
=5m/s^2

v^2-u^2=2as
625-100=10s
s=525/10
=52.5m

Answer 2

Concept introduction:

acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight forward is accelerated when it accelerates or decelerates. Even though the speed is constant, motion on a circle accelerates because the orientation is always shifting.

Given:

Here it is given that the car is accelerated uniformly from $36 km$ per hour to $90 km$ in $3$ seconds.

To find:

We have to find the acceleration of car the distance covered by car in that time.

Solution:

According to the question,

$u=36km/h=10m/s\\v=90km/h=25m/s\\t=3sec$

Now,

$a=(v-u)/t\\=15/3=5m/s^2$

$v^2-u^2=2as\\625-100=10s\\s=525/10\\=52.5m$

Final answer:

Hence, the final answer of the question is => acceleration is $5m/s^{2}$ and distance covered  is $52.5m$.

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