Answer to Question #135035 in Classical Mechanics for Jia Bawa

Question

a train travels in 3 minutes a distance of 3.15 km from rest at one station to another station. It is uniformly accelerated for the first 30 seconds and uniformly retarded for the last 15 seconds, the speed being constant for the remaining time. Find maximum velocity, acceleration and retardation. Use v-t graph to solve the problem.

Answer 1

solution

given data are

total displacement traveled by train(s)=3.15 km

total time taken (t)=3 minute = 180 sec

graph for this equation can be drawn as according to the question

are under the v-t curve will give displacement so

displacement in first 30 sec

$s_1=\frac{1}{2}v_{max}t_1$

by putting the value of t₁

$s_1=\frac{30}{2}v_{max}$ .......eq.1

displacement during constant velocity in 135 sec

$s_2=v_{max}t_2 \\t_2=135s\\so\\s_2=135v_{max}......eq.2$

displacement in last 15 sec

$s_3=\frac{1}{2}v_{max}t_3\\t_3=15s\\so\\s_3=\frac{15}{2}v_{max}......eq.3$

and

$s_1+s_2+s_3=3.15$

from eq.1 ,2 and 3

$\frac{30}{2}v_{max}+135v_{max}+\frac{15}{2}v_{max}=3.15\\and\\\frac{315}{2}v_{max}=3.15\\v_{max}=0.02km/s=20m/s$

and

for initial 30s applying first law of motion

$v_{max}=a_130 \\a_1=\frac{20}{30}=0.66m/s^2$

for last 15 s applying first law of motion

$v_{max}=a_215\\a_2=\frac{20}{15}=1.33m/s^2$

therefor maximum velocity is 20 m/s , acceleration is 0.66 m/s^2 and retardation is 1.33 m/s^2 .