Ask Questions, Get Answers

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

An elevator car whose floor to ceiling distance is 2.7 m starts ascending with constant acceleration of $1.2 m/s^2$ after 2 seconds of the start a bolt begins falling from ceiling of elevator. What is the time after which the bolt hits the floor of the elevator.

\[(a)\;0.8 s\quad (b)\;0.7 s \quad (c)\;0.6s \quad (d)\;0.5 s\]

Can you answer this question?

1 Answer

0 votes
Considering elevator at rest the relative acceleration of bolt is
Now after 2 seconds after the car starts moving with acceleration $1.2 m/s^2$ its velocity will be
$v_{\large car}$$=at=(1.2) \times 2=2.4 m/s$
But w.r.t elevator the bolt has zero velocity , and acceleration $=11 m/s^2$
Therefore $t=\sqrt {\large\frac {2s}{a}}$
$\qquad\qquad=\sqrt {\large\frac{2 \times 2.7}{11}}$$=0.7 s$
Hence b 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, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App