Question

A car of $1400\text{kg}$ is moving on a circular path of radius $30\text{m}$ with a speed of $40\text{km/h}$. When the driver applies the brakes and the car continues to move along the circular path, what is the maximum deceleration possible if the tyres are limited to a total horizontal friction of $10.6\text{kN}$

• A. $10{\text{m/s}}^2$
• B. $6.36{\text{m/s}}^2$
• C. $4{\text{m/s}}^2$
• D. None of these

Answer 1

The correct option is B $6.36{\text{m/s}}^2$

The net acceleration to the car is provided by force of friction
$a=\sqrt{a_c^2+a_t^2}$
$F=ma=m\sqrt{{\left(\frac{v^2}{r}\right)}^2+a_t^2}$
$\Rightarrow 10.6\times {10}^3=m\sqrt{\frac{v^4}{r^2}+a_t^2}$
After solving, $a_t=6.36{\text{m/s}}^2$
Hence, the correct answer is option (b).