Question

A car starts from rest and accelerates uniformly for 10 sec to a velocity of 8 m/s.it then runs at a constant velocity and is finally brought to rest in 64 meters with a constant retardation. The total distance covered by the car is 584 meters. Find the value of acceleration,retardation and total time taken.

Answer 1

Step 1: Given that

Initial velocity($u$ ) of the car= $0$

time for uniform acceleration($a$ )= $10s$

final velocity of the car during acceleration($v$)= $8m{s}^{-1}$

Total distance covered, $d=584m$

Distance travelled during retardation = $64m$

Step 2 : Formula used

$s=ut+\frac{1}{2}a{t}^2$

$v=u+at$

Step 3: Calculation of acceleration

Using first equation of motion, $v=u+at$ , we get

$a=\frac{v-u}{t}$

$a=\frac{8-0}{10}$

$a=0.8m{s}^{-2}$

Step 4: Calculation of distance travelled by the car during this time

Using second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

, we get

$s=0\times 10s+\frac{1}{2}\times 0.8m{s}^{-2}\times {\left(10s\right)}^2$

$s=0.4\times 100$

$s=40m$

Suppose it travels '$x$' distance with constant velocity,

$8m{s}^{-1}$ ,for time 't'.

It then travels 64m with uniform retardation and comes to rest.

Total distance travelled

$40m+x+64m=584m$

$x=584-104$

$x=480m$

So,

$8m{s}^{-1}\times t=480m$

$t=\frac{480}{8}$

$t=60sec$

Let the car travel 64 m with uniform retardation for time $t^{\prime }$

Using,

$v^2=u^2+2as$

$0=8^2+2a\times 64$

$a=-\frac{64}{2\times 64}$

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

retardation =$-0.5m{s}^{-2}$

$t^{\prime }=\frac{0-8}{-0.5}$

$t^{\prime }=16s$

Therefore,
Total time taken is=10s + 60s+16s=86s