Answer to Question #115635 in Mechanics | Relativity for Felix Coleman

Question

1. An electric train accelerates uniformly from rest to a speed of 20 ms−1 which it maintains until the
brakes are applied. It is then brought to rest by a uniform retardation equal in magnitude to twice its
former acceleration. The total distance covered is 7.8 km and the total time taken is 7 minutes. Sketch a
velocity–time diagram. Calculate
(a) the time for which the train is travelling at constant speed,
(b) the initial acceleration in ms−2
.
2.A particle moves so that its position vector r at time t is
r = a cos ωt + b sin ωt,
where ω is a constant and a and b are constant vectors. Show that
(a) r · r˙ is independent of t,
(b) the acceleration is everywhere towards the origin and proportional to r.

Answer 1

1) Let $t_1$ be the time of acceleration, $t_2$ be the time of constant velocity and $t_3$ be the time of deceleration. When the train accelerates or decelerates, its velocity changes linearly. If the module of deceleration is twice the module of acceleration, then the time $t_3 = 0.5t_1$ .

Next, $t_1+t_2+t_3 = 7~\mathrm{min}$ or $t_1+t_2+0.5t_1 = 420\,\mathrm{s}.$

The total distance is the area under $v(t)$ curve. It can be calculated as

$S = 0.5vt_1 + vt_2 + 0.5vt_3 = 0.75vt_1 + vt_2 = 7800\,\mathrm{m}.$

We obtained a linear system

$\begin{cases} 1.5t_1+t_2 = 420,\\ 0.75\cdot20\cdot t_1 + 20t_2=7800. \end{cases}$

Solving it, we get $t_1=40\,\mathrm{s}, \;\; t_2 = 360\,\mathrm{s}.$ Next, acceleration $a = \dfrac{v-0}{t_1} = \dfrac{20}{40}=0.5\,\mathrm{m}/\mathrm{s}^2$ .

So the time of travelling with constant velocity is 360 s and the acceleration is 0.5 m/s² .

2) $\vec{r} = \vec{a}\cos ωt + \vec{b} \sin ωt,$ $\dot{\vec{r}} = -\vec{a}\omega\sin\omega t + \vec{b}\omega\cos\omega t.$

a) We can see that if $t=0$ , then $\;\; \vec{r}=\vec{a}, \;\; \dot{\vec{r}} = \omega\vec{b}, \;\; \vec{r}\cdot\dot{\vec{r}} = \omega\vec{a}\vec{b}$ , but if $\omega t = 90^\circ$ , then $\;\; \vec{r}=\vec{b}, \;\; \dot{\vec{r}} = -\omega\vec{a}, \;\; \vec{r}\cdot\dot{\vec{r}} = -\omega\vec{a}\vec{b}$ , so $\vec{r}\cdot\dot{\vec{r}}$ depends on time.

Let us calculate $\vec{r}\times\dot{\vec{r}}$ :

$\vec{r}\times\dot{\vec{r}} = (\vec{a}\cos ωt + \vec{b} \sin ωt)\times( -\vec{a}\omega\sin\omega t + \vec{b}\omega\cos\omega t) = 0 - \omega \vec{b}\times\vec{a}\sin^2\omega t + \omega\vec{a}\times\vec{b}\cos^2\omega t + 0 = \omega \vec{a}\times\vec{b}.$

We can see that it doesn't depend on time.

b) $\ddot{\vec{r}} = -\vec{a}\omega^2\cos\omega t - \vec{b}\omega^2\sin\omega t = -\omega^2\vec{r}$ , so two vectors are directed oppositely and the acceleration is proportional to $\vec{r}$ with coefficient $-\omega^2.$