# How to solve this variable acceleration Q, parts a to c?

## thank you

Jun 8, 2018

a) $t = 0$, $t = \frac{1}{2}$ and $t = 1$.
b) $\frac{33}{16}$
c) see below

#### Explanation:

The expression for the position can be rewritten in the form :
$x = \frac{1}{2} {t}^{2} {\left(t - 1\right)}^{2}$

a) The velocity is

$v = \frac{\mathrm{dx}}{\mathrm{dt}} = t {\left(t - 1\right)}^{2} + {t}^{2} \left(t - 1\right)$
$\quad = t \left(t - 1\right) \left(2 t - 1\right)$

Thus the particle is at rest ($v = 0$) at the times $t = 0$, $t = \frac{1}{2}$ and $t = 1$.

b) Since the velocity, which in this case is a continuous function, vanishes at $t = 0$, $t = \frac{1}{2}$ and $t = 1$, it must change sign at this points. So, to calculate the distance traveled, we must consider each of these time intervals separately. Now

• displacement between $t = 0$ and $t = \frac{1}{2}$ :
$x \left(\frac{1}{2}\right) - x \left(0\right) = \frac{1}{2} \times {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2} - 1\right)}^{2} - 0$
$q \quad = \frac{1}{32}$
• displacement between $t = \frac{1}{2}$ and $t = 1$ :
$x \left(1\right) - x \left(\frac{1}{2}\right) = 0 - \frac{1}{2} \times {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2} - 1\right)}^{2}$
$q \quad = - \frac{1}{32}$
• displacement between $t = 1$ and $t = 2$ :
$x \left(2\right) - x \left(1\right) = \frac{1}{2} \times {\left(2\right)}^{2} {\left(2 - 1\right)}^{2} - 0$
$q \quad = 2$

The distances traveled in these three intervals are $\frac{1}{32}$, $\frac{1}{32}$ and $2$ respectively.

Thus the total distance traveled is $\frac{1}{32} + \frac{1}{32} + 2 = \frac{33}{16}$

c) It is easy to see that $\frac{1}{2} {t}^{2} {\left(t - 1\right)}^{2}$ can never be negative.