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A PARTICLE STARTS MOVING WITH INITIAL VELOCITY 5m/s.IT IS ACCELERATING WITH A CONSTANT ACCELERATION 2m/sec(square) AT AN ANGLE 120(DEGREE) WITH THE DIRECTION OF INITIAL VELOCITY.FIND THE TIME AFTER WHICH ITS SPEED WOULD BE ROOT(3) TIMES THAT OF THE INITIAL SPEED.

Bharat Makkar , 10 Years ago
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shashi K Sharma

Last Activity: 10 Years ago

Dear Bharat
 
Just assume the axes as X-axis along the initial direction of of motion and Y-axis being perpendicular to it. The provided acceleration has its components as  ax= -1m/s2 and ay=root3m/s2 thus getting vx=(5- t) & vy = root3t
Now the speed at any time is (vx+ vy2)^.5
As per the question demands it should be (root3)5
Putting the values we get a quadratic equation 2t2 – 5t – 25 =0 and hence the feasible value of t=5
 
Hope, I got the question precisely, for there was no diagram
 

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