Question

A particle moving along a straight line with a constant acceleration of $-4m/s^2$ passes through a point $A$ on the line with a velocity of $+8m/s$ at some moment. Find the distance travelled by the particle in $5$ seconds after that moment.

• A. 26 m
• B. 18 m
• C. 9 m
• D. 32 m

Answer 1

Answer: (A)

$u=+8m/s,a=-4m/s^2$

$v=0$

$\Rightarrow 0=8-4t$ or $t=2\sec$.

Displacement in first $2\sec ,$

$S_1=8\times 2+\frac{1}{2}.(-4)\cdot 2^2=8m$

Displacement in next $3\sec ,$

$S_2=0\times 3+\frac{1}{2}(-4)3^2=-18m$

distance travelled $=|S_1|+|S_2|=26m$

ALITER :



We can draw velocity time graph of the situation, total distance is equal to magnitude of area under velocity time graph Hence

$d=\frac{1}{2}\times 2\times 8+\frac{1}{2}\times 3\times 12$

$=8+18=26m$