Question

A car starts from rest and acquires a velocity of 54 km/hr in 2 mins find the acceleration and distance traveled by car during this time assume motion of the car

Answer 1

u=0 v= 54 t=120 s a=? s= ?
Firstly
v=u+at
Then
s=ut+1/2atsquare

Answer 2

Given that,

Intial velocity u=0m/s

Final velocity v=54km/h=54×

185 =15m/s Time t=2min=120s

Now, put the value in equation of motion

The acceleration is v=u+at a= tv−u a= 12015−0a=o.125m/s 2

Now, again from equation of motion

The distance is s=ut+ 21 at 2s=0+ 21 ×0.125×120×120

s=900m

Hence, the acceleration is 0.125m/s 2

and distance is 900m