Question

a car starts from rest and requires 54kn/h in 3 mins find the acceleration and distance traveeled by the car during this time assume motionof a car is uniform ​

Answer 1

Explanation:

Acceleration

Distance

Final Velocity

v = u + a*t

Initial Velocity u

m/s

Final Velocity v

m/s

Time t

s

Acceleration a

m/s²

RESET VALUES

ANSWER

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=12015−0

 a=o.125m/s2

Now, again from equation of motion

The distance is

  s=ut+21at2

 s=0+21×0.125×120×120

 s=900m

Hence, the acceleration is 0.125m/s2 and distance is 900m

Answer 2

Initial velocity=0m/s

Final velocity=54×5/18=15m/s

Time =3mins

Acceleration=v-u/t

a= 15-0/3

a=15/3

a=5m/s^2