Question

The position of a particle as a function of time t, is given by x(t) = at + bt² - ct³ where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:
(A) a+(b²/4c)
(B) a+(b²/c)
(C) a+(b²/2c)
(D) a+(b²/3c)

Answer 1

Answer:

sorry I don't know because I am in class 5

Answer 2

Velocity will be $a+(\frac{b^{2} }{3c} )$ , option D is correct.

Explanation:

Given :

The position as a function of time is given by $x (t) = at + bt^{2}-ct^{3}$.

We know that velocity is the first derivative of displacement wrt. time.

⇒ $v = \frac{dx}{dt}$

   $v = a+2bt-3ct^{2}$

Acceleration is the first derivative of velocity wrt. time.

⇒ $a = \frac{dv}{dt}$

   $a = 2b -6ct$

Given in question when acceleration is zero $(a = 0)$ means,

  $0= 2b -6ct$

 $2b = 6ct$ ; $t = \frac{2b}{6c}$

put $2b$ value in velocity equation,

  $v = a +6ct^{2} -3ct^{2}$ (put value of $t$)

  $v = a +2ct^{2}$

 $v =a + \frac{12cb^{2} }{36c^{2} }$

  $v = a + \frac{b^{2} }{3c}$

So when particle attain zero acceleration then $v = a+(\frac{b^{2} }{3c} )$