Question

a car accelerates from rest with a constant acceleration a on a straight road . after gaining a velocity v, the car moves with that velocity for sometime . then the car decelerates with a retardation b. if the total distance covered by the car is equal to s, find the total time of it"s motion?

Answer 1

square root of 2s/a...
use equation s=ut+1/2at^2

Answer 2

You have to separate the motion into 3 parts:
1) Acceleration - \(s_0=ut_0+\frac{1}{2}at_0 \text{}^2\)
2) Constant velocity - \(s_1=v\times t_1\)
3) Retardation - \(s_2=vt_2 + \frac{1}{2}bt_2 \text{}^2\)

\(s = s_0 + s_1 + s_2\)
\(t = t_0 + t_1 + t_2\)

Answer 3

Also, 'u' is the initial velocity of the car (tip: it accelerates from rest)

Answer 4

v/a +v/b+(2s-av^{2}-bv^{2} )/2v