Question

A Car Is Accerlerated uniformly from 36km per hour to 90km per hour in 3 seconds. Calculate: (a) The Acceleration Of Car (b)The Distance Covered By the car in that time

Answer 1

Initial velocity (u) =36km/h
$= 36 \times \frac{1000}{3600}m \: per \: second$
=10m/s

Final Speed (v)=90 km/h
$= 90 \times \frac{1000}{3600}m \: per \: second$
=25m/s

Time=3 second

Since we know,
$acceleration = \frac{v - u}{t}$
$= \frac{25 - 10}{3} \\ = \frac{15}{3} \\ = 5$
So,the Acceleration=5m/s^2

Now,
Distance travelled S=
$= ut + \frac{1}{2} a {t}^{2}$
$= 10 \times 3 \times \frac{1}{2} \times 5 \times 10 \times 10 \\ = 30 + 250 \\ = 280$
So,distance travelled= 280m.