Question

A car starting from rest acquires a velocity of 36 $\frac{km}{h}$ in 5 s .Calculate

(a) its acceleration .

(b) the distance covered by it .

Answer 1

step 1;

Given that

Initially velocity = 0 .

Finally velocity = 36 $\frac{km}{h}$ in $\frac{m}{s}$ is 36 $\times \frac{5}{18}$=$10\frac{m}{s}.$

Time taken = 5 s.

Step 2:

Acceleration = $\frac{changeinvelocity}{timetaken}=\frac{v-u}{t}.$

Distance = Velocity $\times$time .

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

Step 3 :

(a) Acceleration = .$\frac{changeinvelocity}{timetaken}=\frac{v-u}{t}.$

Acceleration =$\frac{10-0}{5}=2\frac{m}{s^2}.$

(b) Distance covered by it = Velocity $\times$time .

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

$s=0\times 5+\frac{1}{2}\times 2\times 5^2.$

$S=25m.$

Step 4:

Hence, acceleration is $2\frac{m}{s^2}$ and Distance covered = 25 m.