# calculus, velocity, time acceleration

The motion of a body is modelled by the following relationship:

$$s = t^3 - 3t^2 + 3t + 8$$

Where: s = distance in meters, and t = time in seconds

Use calculus to determine the following:

a) the velocity of the body at the end of 3 seconds

b) the time when the body has zero velocity

c) By finding the second derivative of the above relationship the acceleration of the body after 2 seconds

d) when the bodies acceleration is zero

This is my answer to a). I am trying to complete all the questions.. I am trying to learn this. Is my answer correct? If not how have i gone wrong?

$$v(t) = s(t) = t^3 - 3t^2 + 3t + 8$$ $$v(t) = s(t) = 3t^2 - 6t + 3$$ $$v(3) = s(3) = 3(3)^2 - 6(3) + 3$$ $$= 15m/sec$$

• Luke: v=d/dt (s)=3t^2-6t+3, at t=3: v(3)=3(3^2)-6(3)+3. Commented Mar 29, 2020 at 14:05
• I do not think that you correctly differentiate $s$, isnt it suppose to be $v(t)=s'(t)=3t^2-6t+3$. And for the next parts, acceleration is the second derivate of position $s$ wrt to time $t$. Commented Mar 29, 2020 at 14:07
• Yes basic mistake, thanks for noticing. I have edited my answer. Is this correct now?
– Luke
Commented Mar 29, 2020 at 14:11
• No, the notation is still incorrect, $v(t)\ne s(t)$. Commented Mar 29, 2020 at 14:19
• Could someone show me where ive gone wrong?
– Luke
Commented Mar 29, 2020 at 14:23

$$a)$$ $$v(t) = \dot{s}(t) = 3t^2 - 6t + 3$$, so $$v(t = 3) = 3 \cdot 9 - 6 \cdot 3 + 3 = 12$$ (with units $$ms^{-1}$$)

$$b)$$ Solve: $$v(t) = 0$$, which is $$3t^2 - 6t + 3 =0$$ or equivalently $$t^2 - 2t + 1 = 0$$. Factorisation leads to $$(t - 1)^2 = 0$$, so the solution is $$t = 1$$ (with units $$s$$)

$$c)$$ $$a(t) = \ddot{s}(t) = 6t - 6$$, so $$a(t = 2) = 6 \cdot 2 - 6 = 6$$ (with units $$ms^{-2}$$)

$$d)$$ Solve: $$a(t) = 0$$, which is $$6t - 6 = 0$$. So the solution is $$t = 1$$ (with units $$s$$)

Note: I use the dot to denote once differentiating $$s$$ with respect to time $$t$$.

• this is answers to all the questions? I will look at these and understand how to work them out
– Luke
Commented Mar 29, 2020 at 14:54
• Yes. If you find you don't understand any part of what I have done please ask :)
– user486957
Commented Mar 29, 2020 at 14:56
• Thank you i will.
– Luke
Commented Mar 29, 2020 at 14:59
• No worries, good luck.
– user486957
Commented Mar 29, 2020 at 15:22