Question

$\begin{array}{c}\text{A body moves in a circular path of radius R with}\\ \text{deceleration so that at any moment of time its}\\ \text{tangential and normal accelerations are equal in}\\ \text{magnitude. At the initial moment t}=0,\text{the}\\ \text{velocity of body is}{v}_0\text{then the velocity of body}\\ \text{will be-}\end{array}$

• A. $v=\frac{v_0}{1+\left(\frac{v_{0}t}{R}\right)}\text{at time}t$
• B. $v=v_0{e}^{-S/R}\text{after it has moved S meter}$
• C. $v=v_0{e}^{-SR}\text{after it has moved S meter}$
• D. $\text{None of these}$

Answer 1

The correct option is D $\text{None of these}$

Solution we know,

Normal acceleration $\left(a_n\right)=w^{2}R$

Tangent acceleration $\left(a_T\right)=\alpha R$

As

$a_t=a_n$ [given]

$\Rightarrow \alpha R=w^{2}R$

$\Rightarrow \alpha =w^2$

$\Rightarrow \frac{dw}{dt}=w^2$

$\Rightarrow {\int }_{w_0}^w\frac{dw}{w^2}={\int }_0^{t}dt$

$\Rightarrow [\frac{-1}{w}{]}_{w_0}^w=[t{]}_0^t$

we know,

$w=v/R$

$\Rightarrow \frac{-R}{v}+\frac{R}{v_0}=t$

$\Rightarrow v=\frac{v_0}{1-\left(\frac{vot}{R}\right)}$

Option - D is correct