Question

6. A car starts from rest and accelerates uniformly for
10 s to a velocity of 8 ms. It then runs at a constant
velocity and is finally brought to rest in 64 m with a
constant retardation. The total distance covered by
the car is 584 m. Find the value of acceleration,
retardation and total time taken.
(Ans. 0.8 ms 2,0.5 ms-2,86 s)​

Answer 1

Hope this helps

Explanation:

i) u = 0 m/s

t = 10 s

v = 8 m/s

From first law of motion : v = u + at

a = v - u /t

  = 8 /10

  = 0.8 m/$s^{2}$ (acceleration)

ii) Distance covered by the car during retardation = 64 m = S

u = 8 m/s

v = 0 m/s (car comes to rest)

From third law of motion : $v^{2} - u^{2} = 2 a S$

$a = \frac{v^{2} - u^{2} }{2S}$

  = - 64 /128 = - 0.5 m/$s^{2}$ (retardation)

iii) Total distance covered by car = 584 m = S

Time taken by car as it accelerates to constant velocity = 10 s

Time taken by car as it starts decelerating

a = - 0.5 m/$s^{2}$

u = 8 m/s

v = 0 m/s

From first law of motion :

t = v - u / a

  = - 8 / - 0.5

  = 16 s

Distance covered by car during the first 10 s

From second law of motion : $S = ut + \frac{1}{2} at^{2}$

S = 0 + 1/2 x 0.8 x 100 = 40 m

Distance covered by car during its constant velocity travel :

S = 584 - [ 64 + 40 ]

  = 480 m

The car travels this 480 m's with constant velocity of 8 m/s

Velocity = Displacement / Time

Time = Displacement / Velocity

        = 480 /8 = 60 s

Total time taken = 10 + 16 + 60 = 86 s