Question

A particle moving along a straight line with constant acceleration of $-2m/s^2$ has a velocity of $6m/s$ at $t=0$ sec. Find the ratio of distance and displacement at $t=5$ sec.

• A. 13 : 5
• B. 17 : 8
• C. 9 : 5
• D. 11 : 3

Answer 1

The correct option is A 13 : 5

$v=u+at=6-2t$
$v=6-2t$
If $v=0\Rightarrow 6-2t=0\Rightarrow 2t=6\Rightarrow t=3sec$
At $t=5\Rightarrow v=6-2\left(5\right)=-4m/s^2$

Velocity time graph for given situation can be considered as



Distance $=A_1+A_2=\frac{1}{2}\left(6\right)\left(3\right)+\frac{1}{2}\left(2\right)\left(4\right)=13$
Displacement $=A_1-A_2=\frac{1}{2}\left(6\right)\left(3\right)-\frac{1}{2}\left(2\right)\left(4\right)=5$
Threfore, the ratio $=13:5$