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# Suppose in an exceedingly circular motion of a particle , the tangential acceleration of the particle is given by $at\Rightarrow 9m/{{s}^{2}}$. The radius of the circle is 4m. The particle was initially at rest . Find the time after which acceleration of the particle makes an angle of ${{45}^{\circ }}$ with the radial accelerations is –A. $\dfrac{1}{3}s$B. $\dfrac{2}{3}s$C. $\dfrac{4}{3}s$D. $1s$

Last updated date: 19th Jul 2024
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Hint: We know that the radial acceleration is equal to the rate of change in velocity . The change which generates in speed during the non uniform circular motion known as tangential acceleration.

Complete step-by-step solution:
let's draw a diagram of this question to solve this question-

here in this problem , tangential acceleration = $9m/{{s}^{2}}$
Radius =$4m$
from the fig we get triangle ABC ,
\begin{align} & \tan {{45}^{\circ }}\Rightarrow \dfrac{at}{ac} \\ & or,1\Rightarrow \dfrac{9}{ac} \\ & or,ac\Rightarrow 9m/{{s}^{2}} \\ \end{align}
Now , we know that tangential acceleration is
$at\Rightarrow \dfrac{dv}{dt}$
$or,dv\Rightarrow at.dt$
$or,\int_{0}^{v}{dv\Rightarrow \int\limits_{0}^{t}{9dt}}$
$or,v\Rightarrow 9t$ $m/s$
Again , we know that centripetal acceleration $\left( ac \right)=\dfrac{{{v}^{2}}}{r}$
Now , using $(ac)=(at)$ we get

$\dfrac{{{v}^{2}}}{r}\Rightarrow \dfrac{dv}{dt}$
$or,\dfrac{{{(9t)}^{2}}}{4}\Rightarrow 9$
$or,\dfrac{81t}{4}\Rightarrow 9$
$or,{{1}^{2}}\Rightarrow \dfrac{4}{9}$
$or,t\Rightarrow \dfrac{2}{3}\sec$
Therefore, the correct option is B) $\dfrac{2}{3}s$.

Note: During circular motion the speed vector changes its direction at each point on the circle . This suggests that the tangential component of acceleration is always non-zero.