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

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

$x(t)=2.17m+\left(4.8\frac{m}{s^2}\right)t^2-\left(.100\frac{m}{s^6}\right)t^6$

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!

• (i) Find $x'(t)$; (ii) Find the "first" time $a$ when $x'(a)=0$; (iii) Find $x''(t)$ and then $x''(a)$. – André Nicolas Feb 15 '14 at 0:36
• 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 $t=-2$. – André Nicolas Feb 15 '14 at 1:14
• 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 $\ t \ = \ 0 \ .$ – colormegone Feb 15 '14 at 1:20
• @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? – hax0r_n_code Feb 15 '14 at 1:29
• 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 $t=0$. – 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.
• Twice $4.8$ is not $9.8$. And the derivative is not right. – André Nicolas Feb 15 '14 at 1:11