A subway train starts from rest at a station and accelerates at a rate of 1.70m/

Question

A subway train starts from rest at a station and accelerates at a rate of 1.70m/s2 for 14.1s . It runs at constant speed for 70.5s and slows down at a rate of 3.47m/s2 until it stops at the next station. Find the total distance covered.

I thought I started off okay, but then I got lost. I computed that for the first 14.1s the train traveled 169m by using the equation:

x= 1/2(1.70 m/s2)(14.1s)2 -I am unsure if this is wrong.

I don't expect anyone to do all the work for me, just some guidance please. Thank you!

Explanation / Answer

Distance travelled while accelerating=ut+1/2at^2

=0+1/2(1.7)(14.1)^2

=168.99 m

Velocity at the end of 14.1 sec=u+at

=0+(1.7)(14.1)

=23.97 m/s

Distance travelled during constant speed=VelocityXTime

=(23.97)(70.5)

=1689.885 m

Distance travelled while decelerating=v^2-u^2/2a

=(0-23.97^2)/2(-3.47)

=82.79 m

So total distance =168.99+1689.88+82.79

=1941.66 m

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