Question

A truck has an initial velocity of $50\text{m/s}$. On the application of brakes truck stops at $200\text{m}$. The acceleration of the truck will be

• A. $-12.5{\text{ms}}^{-2}$
• B. $24.5{\text{ms}}^{-2}$
• C. $-6.25{\text{ms}}^{-2}$
• D. $6.25{\text{ms}}^{-2}$

Answer 1

The correct option is C $-6.25{\text{ms}}^{-2}$

Given,
$u=50\text{m/s},s=200\text{m},v=0$
From $3^{rd}$ equation of motion, we have
$v^2=u^2+2as$
$0=(50{)}^2+2(a)\times 200$
$-2500=400a$
$\Rightarrow a=-\frac{2500}{400}=-6.25{\text{ms}}^{-2}$
Hence, the acceleration of the truck is $a=-6.25{\text{ms}}^{-2}$