Question

A car is moving with a speed of $30{\text{ms}}^{-1}$ on a circular path of radius $225\text{m}$. If its speed is increasing at the rate of $3{\text{ms}}^{-2}$, the net acceleration of the car is

• A. $4{\text{ms}}^{-2}$
• B. $5{\text{ms}}^{-2}$
• C. $6{\text{ms}}^{-2}$
• D. $2{\text{ms}}^{-2}$

Answer 1

The correct option is B $5{\text{ms}}^{-2}$

Given, $v=30{\text{ms}}^{-1}$, $r=225\text{m}$
We know that the acceleration responsible for increasing the linear velocity in a circular motion is tangential acceleration.
Therefore as per question, $a_t=3{\text{m/s}}^2$
Now, net acceleration
$a_{\text{net}}=\sqrt{a_t^2+a_c^2}$
$\Rightarrow a_{\text{net}}=\sqrt{{\left(\frac{v^2}{r}\right)}^2+a_t^2}$
$\Rightarrow a_{\text{net}}=\sqrt{{\left(\frac{900}{225}\right)}^2+3^2}$
$\Rightarrow a_{\text{net}}=\sqrt{16+9}$
$\Rightarrow a_{\text{net}}=\sqrt{25}$
$\Rightarrow a_{\text{net}}=5{\text{ms}}^{-2}$

hence option B is the correct answer