Question

A car starts from rest and accelerates uniformly at the rate of -1m/s¹ for 5sec. It the maintains rhe velocity for 30 s. then the brakes are applied and the car is retarded to rest in 10 s. Find the maximum velocity attained by the car and the total distance travelled by it.

Answer 1

Acceleration $a={ms}^{-2}$
Time for acceleration, $t_1=5sec$
Initial velocity, $\text{u =0}$
Final velocity v, at t = 5 is
$\text{v = u + a t = 0+(1×5)}=5$
v = 5m/s
Distance travelled is calculated by 2nd equation of motion, $s=ut+\frac{1}{2}a{t}^2$
$S_1=\left(0\right)t+\frac{1}{2}\left(1\right){\left(5\right)}^2=12.5m$
maintain same velocity for 30 s, so accelearation will be zero,
so distance travelled = time × velocity,

$S_2=30\times 5=150m$

Now car retarded in 10 sec.

Final velocity v=0

Initial velocity, $u{=5ms}^{-1}$
v = u + a t

$0=5+a\left(10\right)$

$a=-0.5{ms}^{-2}$

Retardation is, $a=-0.5{ms}^{-2}$

So, distance travelled is calculated by 2nd equation of motion,

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

$S_3=\left(5\right)\left(10\right)+\frac{1}{2}(-0.5){\left(10\right)}^2=25m$

So total distance travelled =

$\text{12.5 +150+25}$

12.5 +150+25

$\text{= 187.5 m}$

= 187.5 m