IIT JEE , AIEEE Forum - Physics , Mechanics - Kinematics Problem-4

prateek.agarwal

Neone plz solve it...and post the solution here...if the image is not visible enlarge by clicking on it...

ANSWERS

s= 20(1 - e^-t/2)

a= -3.75 m/sec²

prakhar.in

dv/ds = -0.5 from fig

dv = -0.5ds

a= -0.5v

at s= 5 v=7.5 by symmetry from fig

so a=3.75

KAB

From the graph we can see that the line passes through two points (20,0) and (0,10)

So the eqn is 2v=20-s

or 2ds/dt=20-s

or ds/(20-s)=dt/2

Integrating we get -ln(20-s)=t/2+c

When t=0 , s=0 so c= -ln20

So ln(20/(20-s))=t/2

or (20-s)/20=e^-t/2

or s=20(1-e^-t/2).......(1)

Now to find acceleration differentiate (1) twice wrt t .Also find t when s=5m from (1)

You will get a = -3.75m/s²

prateek.agarwal

thnx againb KAB...rating u...