Question

A car accelerates uniformly from $13{\text{ms}}^{–1}$ to $31{\text{ms}}^{–1}$ while entering the motorway, covering a distance of $220\text{m}$. Then the acceleration of the car will be:

• A. $2.9{\text{ms}}^{-2}$
• B. $1.8{\text{ms}}^{-2}$
• C. $4{\text{ms}}^{-2}$
• D. $2.2{\text{ms}}^{-2}$

Answer 1

The correct option is B $1.8{\text{ms}}^{-2}$

Given,
$s=220\text{m}$
$u=13{\text{ms}}^{-1}$
$v=31{\text{ms}}^{-1}$
$a=?$
Using $3^{rd}$ equation of motion i.e.
$v^2=u^2+2as$
$(31{)}^2=(13{)}^2+(2\times a\times 220)$
$961=169+440a$
$440a=961-169=792$
$\therefore a=\frac{792}{440}=1.8{\text{ms}}^{-2}$