Question

A car moving with a constant acceleration covers the distance between two points $180m$ apart in $6s$. If its speed as it passes the second point is $45m{s}^{-1}$, its speed at the first point is

• A. 10 $m{s}^{-1}$
• B. 15 $m{s}^{-1}$
• C. 30 $m{s}^{-1}$
• D. 45 $m{s}^{-1}$

Answer 1

Answer: (B)

15 $m{s}^{-1}$

$180=u\left(6\right)+\frac{1}{2}a(6{)}^2$
$180=6u+18a$
$30=u+3a..................\left(1\right)$
$45=u+at$
$45=u+a\left(6\right).................\left(2\right)$
$e{q}^n\left(2\right)-e{q}^n\left(1\right)$
$\Rightarrow 15=3a\Rightarrow a=5$
$u=15m/s$