First slide

Instantaneous acceleration

Question

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

Difficult

A

4 sec
B

8 sec
C

12 sec
D

16 sec

Solution

Given x = bt³ -ct â¦..(1)
from eq .(1), $\frac{dx}{dt} = v = 3 {bt}^{2} - c$ â¦.(2)
and $\frac{dv}{dt} = a = 6 bt$ â¦.(3)
From the given problem,
t=2sec, a=24m/s² and v= -8m/s
From 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 $(âµ x = 0)$ â¦..(6)
From eq.(6), t² =16 or t=Â±4
$â´$ t = 4sec (negative sign is not possible)

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