A subway train starts from rest at a station and accelerates at a rate of 1.50m/

Question

A subway train starts from rest at a station and accelerates at a rate of 1.50m/s2 for 13.9s . It runs at constant speed for 70.4s and slows down at a rate of 3.52m/s2 until it stops at the next station.

Find the total distance covered.

Express your final answer in kilometers.

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.40s , encountering no appreciable air drag.

How tall, in meters, is the building?

How fast is the brick moving just before it reaches the ground?

A rock is thrown vertically upward with a speed of 19.0m/s from the roof of a building that is 50.0mabove the ground. Assume free fall.

Part A

In how many seconds after being thrown does the rock strike the ground?

Part B

What is the speed of the rock just before it strikes the ground?

Explanation / Answer

Train problem:

Part One: positive acceleration

x1-x0=V0t+at2/2 Since the train stars from rest V0=0, and taking x0=0 (the starting point) we have x=at2/2

x1-x0=(1.50)*(13.9)2/2=144.91 m

The final velocity at the end of part one is Vf1=V0+at=0+(1.50)*(13.9)=20.85 m/s

Part two: constant velocity

The train keeps a constant velocity equal to the final velocity at the end of part one, so x2-x1=Vf1t

x2-x1=(20.85)*(70.4)=1467.84 m

Part three: negative acceleration

x3-x2+Vf1t+at2/2

The time elapsed until the train has stopped is calculated from

Vf3=0=Vf1+at2 or t=sqrt(-Vf1/a)

t=sqrt(-20.85/-3.52)=2.43 s

so x3-x2=(20.85)*(2.43)+(-3.52)*(2.43)2/2=40.27 m

The total distance covered is x3-x0=(x1-x0)+(x2-x1)+(x3-x2)=1653.02 m =1.65 km

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