A car starts from rest and moves along the x-axis with constant acceleration 5ms−2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? pls tell

A car starts from rest and moves along the x-axis with constant accele

1.	A car starts from rest and moves along the x-axis with constant acceleration 5ms−2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest? pls tell
Answer» above Question , the following information is given - A car starts from rest and moves along the x-axis with constant ACCELERATION 5ms−2 for 8 seconds.it then continues with constant VELOCITY.We have to find the distance covered by the car in 12 seconds after it STARTED from rest . Solution - From various LAWS of motion , we know that - Acceleration = ∆ V / t => ( v - u ) / t Here , the car started from rest . So , u = 0 It travelled for 8 seconds .Hence , t = 8 seconds .We also have the value of acceleration of the car given as 5 m / s^2 So , 5 = ∆v / 8=> 5 = v / 8 [ as u = 0 ] => v = 40 m / s.Now , Distance travelled by the car in these 8 seconds => S = ut + ( 1 / 2 ) at ^ 2 => S = ( 1 / 2 ) × 5 × 64 => 5 × 32 => 160 m.Now , it travels with a constant velocity of 40m/s for next 4 seconds .So, distance travelled in next 4 seconds = 160 m.Hence total distance travelled by the car in 12 seconds - => 160 m + 160m=> 320 m ......... Answer