Question

A racer drives his bike starting from rest on a circular track of radius $40\text{m}$ and having tangential acceleration of $3{\text{m/s}}^2$. What is the magnitude of net acceleration of the racer $5\text{s}$ after he starts to drive the bike?

• A. $9{\text{m/s}}^2$
• B. $6.37{\text{m/s}}^2$
• C. $5.37{\text{m/s}}^2$
• D. $7.55{\text{m/s}}^2$

Answer 1

The correct option is B $6.37{\text{m/s}}^2$

Tangential acceleration $\left(a_t\right)=\frac{dv}{dt}$
Hence racer's $a_t=3{\text{m/s}}^2$
Applying kinematic equation:
$v=u+a_{t}t$
speed of racer at $t=5\text{s}$:

$v=0+3\times 5=15\text{m/s}$
The centripetal acceleration of racer is given by:
$a_c=\frac{v^2}{r}$
$\Rightarrow a_c=\frac{225}{40}{\text{m/s}}^2$


Hence, net acceleration of racer is given by:
$a_{net}=\sqrt{{a_c}^2+{a_t}^2}=\sqrt{\left(\frac{{225}^2}{1600}\right)+9}=\sqrt{\frac{65025}{1600}}$

$\Rightarrow a_{net}=\sqrt{\frac{(15\times 17{)}^2}{{40}^2}}$

$\therefore a_{net}=\frac{255}{40}=6.375{\text{m/s}}^2$