Question

A car starting from rest accelerates uniformly with 5m/s² for sometime and then decelerates to come to rest with 3m/s² Find the max velocity during covered in a total time of 6 seconds of the journey

Please answer faast

Answer 1

analytical method : Car starting from rest , so initial velocity , u = 0

Let for time t, car is accelerating with, a 5m/s²

then, velocity of car after time t, v = u + at

or, v = 0 + 5t.....(1)

and displacement covered by particle, s = ut + 1/2 at²

= 0 + 1/2 × 5 × t²

= (5/2)t²

after time t, car is decelerating with, b = 3m/s²

Let after time T, car comes to rest.

so, final velocity of car , v = 0

initial velocity in this case, u = 5t

using formula, v = u + at

0 = 5t + (-3)T

or, 5t = 3T.....(2)

given, t + T = 6sec ....(3)

from equations (2) and (3),

t = 18/8 sec , and T = 30/8 sec

so, maximum velocity, v = 5t

= 5 × 18/8 = 90/8 m/s

shortcut : net acceleration, A = ab/(a + b)

and maximum velocity, v = ab(t + T)/(a + b)

applying value of a, b, (t + T)

we get, maximum velocity is v = 90/8 m/s