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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A car exhibits a constant acceleration of $0.300 m/s^2$ parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius $500\; m$. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of $6.00\; m/s$. What is the direction of the total acceleration vector for the car at this instant? 

$\begin{array}{1 1}-13.5 ^{\large\circ}\\-23.7 ^{\large\circ}\\-26.8 ^{\large\circ}\\-11.3 ^{\large\circ}\end{array} $

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1 Answer

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Answer : $-13.5 ^{\large\circ}$
With v = 6.00 m/s and r = 500 m, we find that the radial acceleration is
$a_r=-\large\frac{v^2}{r}$
$\Rightarrow -\large\frac{(6.00m/s)^2}{500m}$$=-0.0720m/s^2$
The radial acceleration vector is directed straight downward while the tangential acceleration vector has magnitude $0.300 m/s^2$ and is horizontal.
Because $a=a_r+a_t$,the magnitude of a is
$a=\sqrt{a_r^2+a_t^2}$
$\Rightarrow \sqrt{(-0.0720)^2+(0.300)^2}m/s^2$
$\Rightarrow 0.309m/s^2$
If $4\pi$ is the angle between a and the horizontal,then
$\phi=\tan^{-1} \large\frac{a_r}{a_t}$
$\Rightarrow \tan^{-1}\big(\large\frac{-0.0720m/s^2}{0.300m/s^2}\big)$
$\Rightarrow -13.5^{\large\circ}$
answered Aug 14, 2014 by sreemathi.v
 

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