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 starts with constant acceleration $ a=2 m/s^2$ at $t=0$ Two coins are released from the car at $t=3$ and $t=4$. Each coin takes 1 second to reach the ground.Then distance between two coins is (assume the coin stick to ground)

\[(a)\;9 m \quad (b)\;7 m \quad (c)\;15 m \quad (d)\;2 m\]

Can you answer this question?

1 Answer

0 votes
Velocity of car at $t=3,v_1=6 m/s$
at $t=4,v_2=8 m/s$
Coin 1 will fall horizontally with velocity 6 m/s
Coin 2 will fall horizontally with velocity 8 m/s
Both coins will travel 6 m and 8 m horizontally before they fall from the point of release.
car moves between t=3 and t=4 $ \large\frac{6 +8}{2} $$\times 1=7 m$ (all the distance are measured from the position at time t=3)
In fourth second, position of first coin $x_1=6$ and position of 2nd coin $= 7+8=15$
Therefore distance between two coins $ =15-6=9 m$
Hence a is the correct answer.
answered Jun 27, 2013 by meena.p
edited May 24, 2014 by lmohan717

Related questions

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