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Displacement (x) of a particle is related to time (t) as x = at + bt 2 – ct 3 where a, b and are constants of motion. The velocity of the particle when its acceleration is zero is given by (A) a + b 2 /c (B) a + b 2 /2c (C) a + b 2 /3c (D) a + b 2 /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.......


Grade:10

3 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

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

Ashwin Muralidharan IIT Madras
290 Points
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).

basit ali
41 Points
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!!!Smile

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