Question

A particle is moving in a circle of radius $r=2\text{m}$. Its speed with time varies according to the relation $v\left(t\right)=4-\frac{6}{t^2}$. What is the magnitude of the total acceleration of the particle at $t=2\text{sec}$ ? Use the value of $\sqrt{769}=27.73$ for calculation.

• A. $2.11{\text{m/s}}^2$
• B. $3.46{\text{m/s}}^2$
• C. $3.1{\text{m/s}}^2$
• D. $1.5{\text{m/s}}^2$

Answer 1

The correct option is B $3.46{\text{m/s}}^2$

Given the variation of the speed of the particle is:
$v\left(t\right)=4-\frac{6}{t^2}$
Diffrentiating both sides w.r.t time, we get:
$\frac{dv}{dt}=0+\frac{12}{t^3}$

Now, tangential acceleration $a_t$ is the rate of change of speed with time,
i.e $a_t=\frac{dv}{dt}=\frac{12}{t^3}$
At $t=2\text{s}$,
$a_t=\frac{12}{8}=\frac{3}{2}{\text{m/s}}^2$

Now speed at $t=2\text{s}$:
$v(t=2)=4-\frac{6}{4}=\frac{5}{2}\text{m/s}$
So, centripetal acceleration at $t=2\text{sec}$
$\therefore a_c=\frac{v^2}{r}=\frac{\left(\frac{25}{4}\right)}{2}=\frac{25}{8}{\text{m/s}}^2$

Total acceleration $a_T$ of particle is
$a_T=\sqrt{a_c^2+a_t^2}=\sqrt{\frac{625}{64}+\frac{9}{4}}=\frac{\sqrt{769}}{8}$
$=\frac{27.73}{8}=3.46{\text{m/s}}^2$