Question

A body leaving a certain point $O$ moves with an acceleration which is constant. At the end of the first second after it left $O$, its velocity is $1.5m/\sec$. At the end of the sixth second after it left $O$ the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

Answer 1

$v=u+at$

$1.5=u+5a...........\left(i\right)$

$0=u+6a.............\left(ii\right)$

From (i) and (ii)

$0=1.5+a(6-5)$

$a=-1.5m/s^2$

Om substituting $a=-1.5m/s^2$ in equation (ii)

$0=u+6(-1.5)$

$0=u-9$

$u=9m/s$

The distance traveled by the body in its forward journey is $s_1$ can be calculated as follows,

$s_1=ut+\frac{1}{2}a{t}^2$

$s_1=9\times 6+\frac{1}{2}\times (-1.5)\times 6^2$

$s_1=54+(-1.5\times 18)$

$s_1=27m$

The distance traveled by the body in its forward journey $\left(s_1\right)$= the distance traveled by the body in its backward journey$\left(s_2\right)=27m$

So, the total distance traveled by the body in forward journey and its backward journey$=s_1+s_2=27+27=54m$