This is a three-part question to which I have already completed the first two but having trouble with how to calculate the third.
Q/ A particle moves in a straight line so that its position $x$ metres at time $t$ seconds is given by $x=t(t-4)^2.$ Find: a) The velocity at time $t.$
$$v(t)=dx/dt =t\times2(t-4)+(t-4)^2 =(t-4)(2t+t-4) =(t-4)(3t-4).$$
b) The value(s) of $t$ when the particle is instantaneously at rest.
$$x(t)=(t-4), t=4,$$ and
$$x(t)=(3t-4), t=4/3.$$
c) What is the acceleration after $4$ seconds?
Part (c) is where I'm having trouble. I have searched through the notes but still unsure on how to calculate. Any help with how to solve would be helpful and appreciated.
Thank you.
The acceleration is the derivative of velocity. So you just differentiate the expression $v(t)$ once and evaluate this when $t=4.$
In addition, I'm not sure I understand what you did in the first two parts. In part (a), you want to differentiate $x(t)$ once to find $v(t).$ In part (b), you want to solve the equation $v(t)=0$ for $t.$