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

Question #115635
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.
Expert's answer

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}\n1.5t_1+t_2 = 420,\\\\\n0.75\\cdot20\\cdot t_1 + 20t_2=7800.\n\\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 \u03c9t + \\vec{b} \\sin \u03c9t," "\\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 \u03c9t + \\vec{b} \\sin \u03c9t)\\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."

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