The position of the front bumper of a test car under microprocessor control is given by:


Find its acceleration at the first instant when the car has zero velocity.

So I'm having trouble understanding where to start. I believe I need to take the derivative of the equation, but how do I find its acceleration when the car has zero velocity?

Any tips to get me started are greatly appreciated!

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    $\begingroup$ (i) Find $x'(t)$; (ii) Find the "first" time $a$ when $x'(a)=0$; (iii) Find $x''(t)$ and then $x''(a)$. $\endgroup$ – André Nicolas Feb 15 '14 at 0:36
  • $\begingroup$ The "first" time when velocity is $0$ is when $9.6 t=0.6 t^5$. This happens at the negative root of $t^4=16$, that is, at $t=-2$. $\endgroup$ – André Nicolas Feb 15 '14 at 1:14
  • $\begingroup$ Although I consider it a sloppy practice, I think it is intended that the position function is implicitly applied for $ \ t \ > \ 0 \ , $ even though it doesn't say so. In that case, we do want the positive fourth-root of 16 (since, otherwise, the first time would still be $ \ t \ = \ 0 \ . $ $\endgroup$ – colormegone Feb 15 '14 at 1:20
  • $\begingroup$ @AndréNicolas the follow up to this question is 'Find its position at the second instant when the car has zero velocity.' How do I find that if the velocity is 0 at only one instant? $\endgroup$ – hax0r_n_code Feb 15 '14 at 1:29
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    $\begingroup$ The equation $x'(t)=0$ has three solutions, $-2$, $0$, and $2$. I do not know whether implicitly $t\ge 0$, answers change if we allow negative $t$. If $t=0$ is first instant, $t=2$ is the second. If $t=-2$ is considered the first instant, the second is $t=0$. $\endgroup$ – André Nicolas Feb 15 '14 at 1:35

$$x(t)=2.17m+\left(4.8\frac{m}{s^2}\right)t^2-\left(.100\frac{m}{s^6}\right)t^6\\ v(t)=\left(9.6\right)t-\left(0.600\right)t^5$$ $v(t)=0$ when $t=0$ (first instant). So, $$a(0)=\left(9.6\frac{m}{s^2}\right)$$ This is indeed the magnitude of the gravitational acceleration.

  • $\begingroup$ I got a similar answer and it says that this is incorrect :( $\endgroup$ – hax0r_n_code Feb 15 '14 at 1:05
  • $\begingroup$ @inquisitor It would be helpful if you were able to hint where I went wrong. $\endgroup$ – user122283 Feb 15 '14 at 1:09
  • $\begingroup$ Twice $4.8$ is not $9.8$. And the derivative is not right. $\endgroup$ – André Nicolas Feb 15 '14 at 1:11
  • $\begingroup$ @AndréNicolas Thanks. The first was a typo, and I corrected the units of the derivative. $\endgroup$ – user122283 Feb 15 '14 at 1:13
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    $\begingroup$ Oh, and there is no reason for the acceleration to have anything to do with Earth surface gravity: it appears simply to be the result of the problem poser's choice of numbers. Presumably, this test car is traveling along the ground... $\endgroup$ – colormegone Feb 15 '14 at 1:24

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