At time t, the velocity of an object is given by the function $v = t^2 – 4t +3$
a) Find the object's acceleration each time the velocity is zero.
b) Find the object's velocity each time the acceleration is zero
for part a) should I evaluate $v(0)$ first? then take a derivative of the function?
Best Answer
for part a) No, you should be finding $t$ when $v(t)=0$, now take the derivative of your velocity function to yield an acceleration function $a(t)$. Use the $t$ or $t$s from your solution to $v(t)=0$ in your acceleration function to calculate the acceleration at that time. Since $a(t)=2t-4$ and since $v(t)=0$ yields $t=1$, $t=3$, it follows that $a(1)=-2$ and $a(3) = 2$. In total, we have found that the velocity equals zero when t = 1 and t = 3. Using that, and the derivative of the velocity function, we found that the accelerations corresponding to those t values are $-2$ and $2$ respectively.
part b) Find the objects velocity each time the velocity is zero? Are you sure you don't mean: find the velocity each time the acceleration is zero? The first question would be trivial so I will assume you mean the latter.
by part a, $a(t) = 2t-4$. Thus when $t=2$, $a(t) = 0$. Now, $v(2) = -1$. So, when the acceleration is 0 the velocity is -1.