Question

A car accelerates uniformly from 18 km h to 36km h in 5 sec calculate acceleration and distance cover by car in that time

Answer 1

Here's your answer...

Solution:-
Change in velocity=(36km/h - 18km/h)
$= > 18kmph$
To convert it in m/s we will multiple it by 5/18
$= > 18 \times \frac{5}{18}$
=>5m/s

Time =5 seconds

We know that,
$acceleration = \frac{change \: in \: velocity}{time} \\ = > \frac{5}{5} = 1$
So,
Acceleration is 1m/s^2

Now,
We will find Distance travelled,by using equation,that is:-
$= > s = ut + \frac{1}{2}a {t}^{2} \\ = > s = 5 \times 5 + \frac{1}{2} \times 1 \times {(5)}^{2} \\ = > s = 25 + \frac{1}{2} \times 25$
$= > s = 25 + 12.5 \\ = >s = 37.5m$
Hence,
Distance travelled is 37.5m

Hope it helps you....