Answer to Question #95763 in Mechanics | Relativity for Musah fuseini

Question

A train starts from rest at station A and is uniformly accelerated until it reaches speed of 108km/h. It then travel at this speed until the breaks are applied and the train is then uniformly retarded until it stops at station B. The magnitude of this retardation is twice the magnitude of the initial acceleration. The distance between the stations is 12km and the time taken for the journey is 10minutes. Find the 1. Time spent of each of the three stages of the journey. 2. Initial acceleration 3. Final retardation. Question 2 A stone 0.2kg is dropped over a cliff. Write down the force in Ns which acts on it and hence find its momentum after 3s. Take g=10m/s

Answer 1

Firstly convert all unit into standard unit

$108\ km\ h^{-1}=30\ m\ s^{-1}\\12\ km=12000\ m\\10 \ min=600\ s$

$Let \ initial\ accelration\ be\ a\\final \ retardation \ be\ 2a\\t_1=time\ for\ accelration\\t_2=time for constant\ velocity\\t_3=time\ for\ retardation$

$v_1=at_1\\v_1=2at_3\\v_1=30\\t_1=time \ elapsed\ for \ acceleration\\total\ distance =s_1+s_2+s_3\\s_1=\frac{1}{2}at_1^2\ \ \ \ \ \ \ \ s_2=v_1t_2\\s_3=at_3^2$

$S=\frac{v_1t_1}{2}+v_1t_2+v_1t_3\\S=15t_1+30t_2+15t_3=12000\\also \ \ \ t_1=2t_3\\so,\ \ \ 30t_2+45t_3=12000\ \ \ \ ........(1)\\also,\ \ t_1+t_2+t_3=600\ \ \ \ \ \\t_2+3t_3=600\ \ \ \ ........(2)\\$

$multiply\ equation \ (2)\ by \ 30 \ and \ subtract\ it \ from (1)$

$45t_3=6000\\t_3=133.33\ s\\t_1=2t_3=266.66\ s\\t_2=600-3t_3=600-400=200 \ s$

$(1) time \ elapsed \ of \ each \ stage\ is\ \\ 266.66\ ,200\ and\ 133.3 \ seconds \ respectively$

$(2)$ $v_1=30=at_1=266.66a$

$\ \ \ \ \ \ \ a=\frac{30}{266.66}=0.1125 \ m \ s^{-2}$

$(3)\ Final\ retardation\ =2a=2\times0.1125=0.225\ m\ s^{-1}$

$Ques.\ 2\\m=0.2\ kg\\g=10\\F=mg=0.2\times10=2\ N\\velocity\ attained\ after\ 3\ s=v =g\times3=30\ m\ s^{-1}\\Momentum = mv=0.2\times30=6\ N\ m\ s^{-1}$