Question

A particle moves on a circle of radius $r$ with centripetal acceleration as function of time as $a_e=k^{2}r{t}^2$ where $k$ is a positive constant. Find the resultant acceleration.

• A. $k{t}^2$
• B. $kr$
• C. $kr\sqrt{k^2{r}^4+1}$
• D. $kr\sqrt{k^2{r}^4-1}$

Answer 1

Answer: (B)

$kr$

The centripetal acceleration is given as

$\frac{v^2}{r}=k^{2}r{t}^2$

$v^2=k^2{r}^2{t}^2$

$v=krt$

The acceleration is given by differentiating the above equation

$\frac{dv}{dt}=kr$

$a=kr$