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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?

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

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  • $\begingroup$ Luke: v=d/dt (s)=3t^2-6t+3, at t=3: v(3)=3(3^2)-6(3)+3. $\endgroup$ Commented Mar 29, 2020 at 14:05
  • $\begingroup$ 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 $t$. $\endgroup$
    – sentheta
    Commented Mar 29, 2020 at 14:07
  • $\begingroup$ Yes basic mistake, thanks for noticing. I have edited my answer. Is this correct now? $\endgroup$
    – Luke
    Commented Mar 29, 2020 at 14:11
  • $\begingroup$ No, the notation is still incorrect, $v(t)\ne s(t)$. $\endgroup$ Commented Mar 29, 2020 at 14:19
  • $\begingroup$ Could someone show me where ive gone wrong? $\endgroup$
    – Luke
    Commented Mar 29, 2020 at 14:23

1 Answer 1

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$a)$ $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 $t$.

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

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