Question

a car start from rest and moves along with X-axis with constant acceleration 5m/s² for 8second. if it then continues with constant velocity. what distance will car cover in 12 second since its started from rest

Answer 1

Given, the car starts from rest, so its initial velocity u = 0
Acceleration, (a) = 5 ms-2 and time (t) = 8 s
From first equation of motion,
v = u + at
On putting a = 5 ms-2and t = 8 s in above equation, we get
v = 0+ 5×8= 40ms’1
So, final velocity v is 40 ms’1.
Again, from second equation of motion,



So, the distance covered in 8 s is 160 m. -­
Given, total time t = 12 s.
After 8 s, the car continues with constant velocity i.e., the car will move with a velocity of 40 ms-1.
So, remaining time t’= 12 s-8 s= 4s
The distance covered in the last 4s (s’) =Velocity x Time [∴ Distance = Velocity x Time]
= 40 x 4 = 160 m
[We have used the direct formula because after 8 s, car is moving with constant velocity i.e., zero acceleration].
Total distance travelled in 12 s from the start
D = s + s’= 160+ 160= 320 m