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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 + b2/4cPlease help as I am facing problem with this question.......

## 3 Answers

9 years ago

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).

9 years ago

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*t

=    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!!! ## ASK QUESTION

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