Question

A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 $m{s}^{-1}$. It then runs at a constant velocity and is finally brought to rest in 64 m with constant retardation. The total distance covered by the car is 584 m. Find the value of acceleration, retardation, and total time is taken.

Answer 1

(Refer to Image)

$V=at$

$\Rightarrow 8=a10$

$\Rightarrow a=0.8m/s^2$

$V^2=U^2+2as$

$\Rightarrow U=8^2+2\times a64$

$\Rightarrow a=\frac{-64}{2\times 64}=-0.5m/s^2$

Distance travelled= Area of graph

$\Rightarrow 584=\frac{1}{2}\times 10\times 8+8\times t^1+\frac{1}{2}\times 16\times 8$

$\Rightarrow 584=40+8t^1+64$

$\Rightarrow 8t^1=480$

$\Rightarrow t^1=60$

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