Question

A particle moving along a straight line with a constant acceleration of $-4 \,m/s^2$ passes through a point $A$ on the line with a velocity of $+8 \,m/s$ 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)

$u = +8 \,m/s, a = -4 \,m/s^2$
$v = 0$
$\Rightarrow 0 = 8 - 4 t$ or $t = 2 \, \sec$.
Displacement in first $2\, \sec,$
$S_1 = 8 \times 2 + \frac{1}{2} . (-4) \cdot 2^2 = 8\,m$
Displacement in next $3\, \sec,$
$S_2 = 0 \times 3 + \frac{1}{2}(-4) 3^2 = - 18\,m$
distance travelled $ = |S_1| + |S_2| = 26\,m$
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 = 26\,m$