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Displacement (x) of a particle is related to time (t) as x = at + bt2 – ct3 where a, b and are constants of motion. The velocity of the particle when its acceleration is zero is given by (A) a + b2/c (B) a + b2/2c (C) a + b2/3c (D) a + b2/4c Please help as I am facing problem with this question.......
Displacement (x) of a particle is related to time (t) as x = at + bt2 – ct3 where a, b and are constants of motion. The velocity of the particle when its acceleration is zero is given by
(A) a + b2/c
(B) a + b2/2c
(C) a + b2/3c
(D) a + b2/4c
Please help as I am facing problem with this question.......
velocity=dx/dt=a+2bt-3ct2
acceleration =dv/dt=2b-6ct,
when a=0,t=b/3c
velocity=a+2bt-3ct2,where t=b/3c
Hi Akshay,
The velocity is rate of change of displacement.
And Acceleration is rate of change of velocity.
So v = dx/dt = a+2bt-3ct2.
And acc = dv/dt = 2b-6ct.
So when acc = 0, we have 2b=6ct or t=b/3c.
At this time, the velocity v = a+2b(b/3c)-3c(b/3c)2 = a + b2/3c.
And hence option (C).
All the best.
Regards,
Ashwin (IIT Madras).
x=at+bt2-ct3,
then, v=a+2bt-3ct2(at given situation)
a=2b-6ct=0
2b=6ct .......(1)
v= a+(6ct)t - 3 ct2
= a+6ct2-3ct2
= a+3ct2 =a+ 3*b/3t*t2
= a+ 3ct * t
= a+ bt
= a+ 2b2/6c
= a+ b2/3c
HENCE OPTION (C) IS THE CORRECT ANSWER!!! IF U LIKE IT PLZ APPROVE IT!!!
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