Answer to Question #176394 in Physics for dawit

Question

A train travelling along a straight track starts from rest at point

A and accelerates uniformly to 20 m s–1 in 20 s. It travels at

this speed for 60 s, then slows down uniformly to rest in 40 s

at point C. It stays at rest at C for 30 s, then reverses direction,

accelerating uniformly to 10 m s–1 in 10 s. It travels at this speed

for 30 s, then slows down uniformly to rest in 10 s when it

reaches point B.

a Plot a graph of the motion of the train.

b Use your graph to calculate:

i the train’s displacement from point A when it reaches

point C

ii the train’s displacement from point A when it reaches

point B

iii the train’s acceleration each time its speed changes.

Answer 1

a)

b) i)

$$s_1=0.5(60+120)(20)=1800\ m\\$$

ii)

$$s_2=1800-0.5(30+50)(20)=1400\ m$$

iii) From 0 s to 20 s

$$a=\frac{20-0}{20}=1\frac{m}{s^2}$$

From 80 s to 120 s

$$a=\frac{0-20}{40}=-0.5\frac{m}{s^2}$$

From 150 s to 160 s

$$a=\frac{-10-0}{10}=-1\frac{m}{s^2}$$

From 190 s to 200 s

$$a=\frac{20-(-10)}{10}=1\frac{m}{s^2}$$