Q. A cyclist is travelling on a circular section of highway of radius 2500 ft at the speed of 60 mile $h^{- 1}$. The cyclist suddenly applies the brakes causing the bicycle to slown sown at constant rate. Knowing that after 8 s the speed has been reduced to $45\ mileh^{- 1}$. The acceleration of the bicycle immediately after the brakes have been applied is
A) $2\ {ft/s}^{2}$
B) $4.14\ {ft/s}^{2}$
C) $3.10\ {ft/s}^{2}$
D) $2.75\ {ft/s}^{2}$
Explanation
(b) $Tangential\ acceleration,\ \ a_{t} = \frac{v_{f} - v_{i}}{t}$
$= \left( \frac{45 - 60}{8} \right)\frac{22}{15}\ {fts}^{- 1}$
$= - \frac{11}{4}\ {fts}^{- 2}$
$Radial\ acceleration,\ a_{r} = \frac{v^{2}}{r} = \frac{\left( 60 \times \frac{22}{15} \right)^{2}}{2500} = 3.1\ {fts}^{- 2}\ $
$Then,\ \ a = \sqrt{a_{r}^{2} + a_{t}^{2}} = 4.14\ {fts}^{- 2}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.