Question

Solve the following:

A) An engine is moving with a velocity 44 m/s. After applying the brakes, it stops after covering a distance of 121 m. Calculate retardation and time taken by the engine to stop.

B) A body starts from rest and moves with a constant acceleration. It travels a distance $S_1$ in first 10s, and a distance$S_2$ in next 10s. Find the relation between $S_1$ and $S_2$. [5 MARKS]

Answer 1

Part A : 2 Marks
Part B : 3 Marks

A)
Here,
$u=44m/s,S=121m,v=0$
$Using,v^2=u^2+2aS$
$0={44}^2+(2\times 121\times a)$
$a=-8m/s^2$

Also,
$v=u+at$
$0=44-8t$
$t=5.5s$

B)
Given :
$u=0,t_1=10s$
Let the acceleration be a.
Distance travelled in first 10 seconds, is given by
$S_1=ut+\frac{a{t}^2}{2}$
= $0+\frac{a\times 10\times 10}{2}$
$S_1=50a$ ------- (1)

To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
$S=ut+\frac{a{t}^2}{2}$
= $0+\frac{a\times 20\times 20}{2}$
= $200a$

Distance travelled in the next 10s interval,
$S_2=S–S_1=200a–50a$
$S_2=150a$

Using equation 1,
$S_2=3S_1$