Question

The position of a particle as a function of time t, is given by $$x(t)=at+bt^2-ct^3$$ where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be?

Answer 1

Correct option is D. $$a+\dfrac{b^2}{3c}$$
$$x=at+bt^2-ct^3$$
$$v=\dfrac{dx}{dt}=a+2bt-3ct^2$$
$$a=\dfrac{dv}{dt}=2b-6ct=0\Rightarrow t=\dfrac{b}{3c}$$
$$v_{\left(at t =\dfrac{b}{3c}\right)}=a+2b\left(\dfrac{b}{3c}\right)-3c\left(\dfrac{b}{3c}\right)$$
$$=a+\dfrac{b^2}{3c}$$.