Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
Answer : $-13.5 ^{\large\circ}$
With v = 6.00 m/s and r = 500 m, we find that the radial acceleration is
$\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
$\Rightarrow \sqrt{(-0.0720)^2+(0.300)^2}m/s^2$
$\Rightarrow 0.309m/s^2$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App