Ask Questions, Get Answers

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

A car is going at a speed of 6m/s when it encouters a 15 m slope of angle $30 ^{\circ}$. The frictional coefficient between road and tyre is 0.5. The driver applies the brakes. The minimum speed of car with which he can reach the bottom is $(g=10 m/s^2)$

\[(a)\;4\;m/s \quad (b)\; 3\;m/s \quad (c)\; 8.45 \;m/s \quad (d)\;7.46\;m/s \]

Can you answer this question?

1 Answer

0 votes
acceleration down the slope
$a=g \sin \theta - \mu g \cos \theta$
$a=10 \sin 30-(0.5)(10) cos 30$
$a=0.67 m/s^2$
Minimum speed while reaching the bottom
$v=7.46 m/s$
Hence d is the correct answer


answered Jul 10, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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