Answer to Question #155479 in Mechanics | Relativity for Aisha

Question

An electric train moves from rest with a uniform acceleration of 1.5 for the first 10s and continues accelerating at 0.5 for a further 20s. It continues at a constant speed for 90s and finally takes 30s to decelerate uniformly to rest.

Using the equations of motion calculate the total distance travelled

Answer 1

For the motion with a constant acceleration the distance is obtained as

$S_1 = v_0t + \dfrac{at^2}{2}$ .

For the first 10 seconds the distance will be

$0\cdot10\,\mathrm{s} + \dfrac{1.5\cdot10^2}{2} = 75\,\mathrm{m}.$

The final velocity will be $V_1=V_0 + at = 0 + 1.5\cdot10 = 15\,\mathrm{m/s}.$

For the second interval:

$S_2 = 15\cdot20 + \dfrac{0.5\cdot20^2}{2} = 400\,\mathrm{m}.$

The final velocity will be

$V_2 = 15 + 0.5\cdot20 = 25\,\mathrm{m/s}.$

For the third interval the distance $S_3 = 25\cdot 90 = 2250\,\mathrm{m}.$

The deceleration is $a_3 = \dfrac{0-V_2}{t} = \dfrac{-25}{30} \approx -0.833\,\mathrm{m/s^2}.$

The distance will be

$S_4 = 25\cdot30 + \dfrac{-0.833\cdot30^2}{2} = 375\,\mathrm{m}.$

So the total distance will be

$75+400+2250+375 = 3100\,\mathrm{m}.$