Question

Two bodies, initially separated by ${}^{\prime }{x}^{\prime }$ m and having an initial velocity $2m{s}^{-1}$ each, are moving towards each other along a straight line. If the rate of increase in their speeds is $3m{s}^{-2}$ and $2m{s}^{-2}$ respectively and the two meet after 4 seconds, find $x.$

Answer 1

Given:

For body 1
Initial velocity $u=2m{s}^{-1}$
Acceleration $a_1=3m{s}^{-1}$

For body 2
Initial velocity $u=2m{s}^{-1}$
Acceleration $a_2=2m{s}^{-1}$

They both meet after time $t=4s$

If body 1 moves a distance $s$ then body 2 moves (x-s) distance for $t=4s$.

Applying second equation of motion for body 1

$s=ut+\frac{1}{2}{a}_1{t}^2$
$\Rightarrow s=\left(2\right)\left(4\right)+\frac{1}{2}\left(3\right)(4{)}^2$
$\Rightarrow s=8+24=32m$

Applying second equation of motion for body 2
$s=ut+\frac{1}{2}{a}_2{t}^2$

$(s-x)=\left(2\right)\left(4\right)+\frac{1}{2}\left(2\right)(4{)}^2$
$\Rightarrow (s-x)=8+16=24m$

$\therefore$ Total distance $\left(x\right)=s+(s-x)$
$=32+24=56m$