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

A 2m wide truck is moving with a uniform speed of $v_0=8 m/s$ along a staright horizontal road. A pedestrian starts to cross the road with a uniform speed of v when the truck is 4m away from him. The minimum value of 'v' so that he can cross the road safely is

\[(a)\; 2.62 m/s \quad (b)\;4.6 m/s \quad (c)\;3.57 m/s \quad (d)1.414 m/s \]


Can you answer this question?

1 Answer

0 votes
Let the man crossing the road at an angle $\theta$. For safe crossing the man must cross the road by the time the truck describes a distance $4+AC$
or $4+2 \cot \theta$
$=\large\frac{4+2 \cot \theta}{8}=\frac{2 /\sin \theta}{v}$
$v= \large\frac{8}{2 \sin \theta+\cos \theta}$
for minimum time
$\large\frac{dv}{d\theta}$$=0=>\large\frac{-8(2 \cos \theta- \sin \theta)}{(2 \sin \theta+\cos \theta)^2}$$=0$
$2 \cos \theta-\sin \theta=0$
$\tan \theta=2$
$=>\sin \theta= \large\frac{2}{\sqrt 5}$
$=>\cos \theta= \large\frac{1}{\sqrt 5}$
Therefore $v= \large\frac{8}{2 \times \Large\frac{2}{\sqrt 5}-\frac{1}{\sqrt 5}}$
$=\large\frac{8}{\sqrt 5}$
$=3.57 m/s$
Hence c is the correct answer.
answered Jun 25, 2013 by meena.p
edited May 23, 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