Question

A particle with deceleration along the circle of radius $R$ so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment $t = 0$, the speed of the particle equals ${v_0}$ then the speed of the particle as a function of the distance covered $s$ will be
(A) $v = {v_0}2{e^{ - {s}/{R}}}$
(B) $v = {v_0}{e^{{s}/{R}}}$
(C) $y = {v_0}{e^{ - {R}/{s}}}$
(D) $v = {v_0}{e^{{R}/{s}}}$

Answer 1

Hint
We know that acceleration is defined as the rate of the change in velocity of an object with respect to the time. The orientation of an object’s acceleration is given by the placement of the net force acting on that object. Based on this concept we have to answer this question.

Complete step-by-step answer
It given that: $\dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{r}$
So now we have to evaluate the expression to get:
$\dfrac{{dv}}{{ds}} = \dfrac{{{v^2}}}{r}$
Now,
$- \int\limits_{{v_0}}^v {\dfrac{1}{v}dv} = \int\limits_0^s {\dfrac{{ds}}{r}}$
$\Rightarrow \ln \left[ {\dfrac{{{v_0}}}{v}} \right] = \dfrac{S}{r}$
$\Rightarrow \dfrac{{{v_0}}}{v} = {e^{S/r}}$
So, now we can write that:
$\Rightarrow {v_o} = v{e^{S/r}} \Rightarrow v = {v_0}{e^{ - S/r}}$
Hence, the correct answer is Option (A).

Note
We know that the concept of tangential acceleration is used to measure the changes in the time tangential velocity of the point with a specific radius that changes with the time. The formula of the tangential acceleration is given as the multiplication between radius of the rotation and angular acceleration.