Question

A particle is moving on a circular path of radius $20\text{m}$. The speed of particle at any instant is given by the relation $v=3t^2-5t$ $\text{m/s}$. What is the magnitude of total acceleration of this particle at $t=5\text{s}$ ?

• A. $\sqrt{16250}{\text{m/s}}^2$
• B. $\sqrt{15000}{\text{m/s}}^2$
• C. $\sqrt{6000}{\text{m/s}}^2$
• D. $\sqrt{10500}{\text{m/s}}^2$

Answer 1

The correct option is A $\sqrt{16250}{\text{m/s}}^2$

The equation of speed is given by $v=3t^2-5t$
Tangential acceleration $a_t$ represents rate of change of speed along the circular path:
$\therefore$ tangential acceleration at $t=5\text{s}$
$a_t=\frac{dv}{dt}=6t-5\Rightarrow a_t=6\times 5-5=25m/s^2$

Speed at $t=5\text{s}$ is $v=3\left(5^2\right)-5\left(5\right)=50\text{m/s}$

Radial acceleration or centripetal acceleration will always acts towards the centre of circular path.

$a_c=\frac{v^2}{r}=\frac{2500}{20}\Rightarrow a_c=125m/s^2$

Total acceleration will be found by the vectorial addition:


Magnitude of total acceleration $a_T$ is given by:
$a_T=\sqrt{{a_t}^2+{a_r}^2}=\sqrt{{25}^2+{125}^2}\Rightarrow a_T=\sqrt{16250}m/s^2$