Question

A particle moves in a circle of radius $1.0cm$ at a speed given be $v=2.0t$ where $v$ is in $cm/s$ and $t$ is in seconds. Find the magnitude of net acceleration at $t=1s$.

• A. $4cm/s^2$
• B. $2cm/s^2$
• C. $\sqrt{20}cm/s^2$
• D. $\sqrt{10}cm/s^2$

Answer 1

The correct option is C $\sqrt{20}cm/s^2$

Given,
$v=2.0t$
$r=1.0cm$
As we know,
$a$) Radial acceleration ($a_r$)=$\frac{v^2}{r}$
(Radial acceleration does not depends upon the instantaneous velocity and will be constant throughout the motion because radial acceleration is perpendicular to the tangential velocity)
Velocity $v$ ($t=1s$) = $2\times 1=2cm/s$
hence, radial acceleration ($a_r$)=$\frac{v^2}{r}=\frac{2^2}{1}=4cm/s^2$

$b$) Tangential acceleration ($a_t$)
(tangential acceleration will be in the direction of the instantaneous velocity)
$v=2t$
Differentiate both side w.r.t. time, hence
$\frac{dv}{dt}=2$
Tangential acceleration $a_t=\frac{dv}{dt}=2cm/s^2$

Magnitude of net acceleration ($a_T$) $a_T=\sqrt{\left(a_r^2\right)+(a_t{)}^2}$
$a_T=\sqrt{(4{)}^2+(2{)}^2}$
$a_T=\sqrt{20}cm/s^2$