The position of a point which moves in a straight line is given by x = bt3 - ct where x is in metre, t in second and b and c are positive constants. when t = 2 second, the acceleration is 24 m/s2 in positive x direction and at the same time, the velocity is 8 m/sec in negative x direction. Find the total time t required for the point to return to the origin at x = 0
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a
4 sec
b
8 sec
c
12 sec
d
16 sec
answer is A.
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Detailed Solution
Given x = bt3 -ct â¦..(1)from eq .(1), dxdt=v=3bt2-câ¦.(2)and dvdt=a=6btâ¦.(3)From the given problem,t=2sec, a=24m/s2 and v= -8m/sFrom eq.(3),24=6bx2 or b =2â¦.(4)From eq. (2), - 8 = 3 x 2x 4-c c-24+B=32â¦.(5)From eq.(1), 0 = 2xt3 -32t (âµx=0) â¦..(6)From eq.(6), t2 =16 or t=±4â´ t = 4sec (negative sign is not possible)
The position of a point which moves in a straight line is given by x = bt3 - ct where x is in metre, t in second and b and c are positive constants. when t = 2 second, the acceleration is 24 m/s2 in positive x direction and at the same time, the velocity is 8 m/sec in negative x direction. Find the total time t required for the point to return to the origin at x = 0