Question

Two subway stops are separated by 1.1 km. If a subway train accelerates at +1.2 m/s2 from rest

through the first half of the distance and decelerates at −1.2 m/s2 through the second half, what are
(a) its travel time and (b) its maximum speed? (c) Graph x, v and a versus t for the trip.

Answer 1

a) ttotal=60.58 s

b) v^2=36.3 m/s

c) the attached photo shows the graphics.

Step-by-step explanation:

a) we have the following:

S1=1.1x10^-3/2=550 m

a=1.2 m/s

we have the following formula:

v^2=u^2+2as

replacing values:

v^2=0+(2*1.2*550)=36.33 m/s

The time is equal to:

t1=v/a=36.33/1.2=30.3 s

we have the following:

u=36.33 m/s

v=0

S=550 m/s

using the following expression:

v=u+at2

clearing t2:

t2=(v-u)/a=(0-36.33)/-1.2=30.28 s

ttotal=t1+t2=30.3+30.28=60.58 s

b) the mass speed is equal to:

v^2=u^2+2as=36.3 m/s

Answer 2

a) Travel time = 60.56 s

b) Maximum speed = 36.33 m/s

Step-by-step explanation:

a) Distance = 1.1 km = 1100 m

A subway train accelerates at +1.2 m/s² from rest through the first half of the distance and decelerates at −1.2 m/s² through the second half.

So half the distance is traveled at an acceleration of +1.2 m/s².

We have equation of motion
`s=ut+(1)/(2)at^2`

Substituting

`(1100)/(2)=0* t+(1)/(2)* 1.2t^2\\\\t=30.28s`

Travel time = 2 x 30.28 = 60.56 s

b) We have equation of motion v = u+at

Substituting t = 30.28 s and a = 1.2 m/s²

v = 0 + 1.2 x 30.28 = 36.33 m/s

Maximum speed = 36.33 m/s

c) Photos of graphs are given