Question

A scooterist is approaching a circular turn of radius $80\text{m}$. He reduced his speed from $27{\text{kmh}}^{-1}$ at constant rate of ${\text{0.5 ms}}^{-2}$. Find net acceleration and angle made by it wrt tangential deceleration

• A. ${\text{0.86 ms}}^{-2},{54}^{\circ }$
• B. ${\text{0.68 ms}}^{-2},{45}^{\circ }$
• C. ${\text{1.0 ms}}^{-2},{45}^{\circ }$
• D. ${\text{0.5 ms}}^{-2},{45}^{\circ }$

Answer 1

The correct option is A ${\text{0.86 ms}}^{-2},{54}^{\circ }$


Centripetal acceleration
$a_r=\frac{v^2}{r}=\frac{{\left(\frac{27\times 1000}{3600}\right)}^2}{80}$
$=\frac{7.5\times 7.5}{80}=0.703{\text{ms}}^{-2}$
Net acceleration, $a=\sqrt{a_r^2+a_t^2}$
where $a_t$ = tangential deceleration $=0.5m{s}^{-2}$

$\therefore a=\sqrt{{0.703}^2+{0.5}^2}=0.86m{s}^{-2}$
and
$\theta =ta{n}^{-1}\frac{a_r}{a_t}=ta{n}^{-1}\frac{0.703}{0.5}={54}^{\circ }$