Question

A car start from rest has an acc

0.3 m/s², calculate the velocity and distance
travelled by this car after 2 min
-​

Answer 1

Velocity of car after 2mins is 36m/s

Distance travelled after 2mins is 2160m

Step-by-step explanation:

A car start from rest with acceleration a=0.3
`(m)/(s^(2) )`

Using equation of motion,

Velocity v is given by :
`v=u+at`

Distance d is given by
`d=ut+(a)/(2)t^(2)`

Velocity after time t=2mins=120s is

`v=u+at`

As car start from rest u=0

`v=(0.3)(120)`

`v=(0.3)(120)`

`v=36m/s`

Therefore, Velocity of car after 2mins is 36m/s

Distance travelled after 2mins=120s

`d=ut+(a)/(2)t^(2)`

As car start from rest u=0

`d=0+(0.3)/(2)(120)^(2)`

`d=2160m`

Therefore, Distance travelled after 2mins is 2160m