course PHY231
Right, and that will give you double what you got for that part of the distance.
Dillon,
Your solution is good except that during the second phase of motion the velocity is constant. The initial velocity is equal to the final velocity, which is in turn equal to the average velocity. There is no justification for averaging velocity zero with your 22.4 m/s velocity on this part of the interval. This doubles the distance on that interval.
Please submit a copy of this document, including my response, using the Submit Work form so I can post this to your access page.
David A. Smith
Associate Professor of Mathematics
Virginia Highlands Community College
dsmith@vhcc.edu
________________________________
From: Dillon Breeding [mailto:dwb2199@email.vccs.edu]
Sent: Wed 9/10/2008 5:55 PM
To: David Smith
Subject: Physics question
""A subway train starts from rest at a station and accelerates
at a rate of 1.60m/s^2 for 14.0s. It runs at a constant speed
for 70.0s and slows down at a rate of 3.50 m/s^2 until it
stops at the next station. Find the total distance covered.
I figured this would be similar to the problem I asked you
early, but since these is an even numbered problems the
answers aren't in the back of the book.
Anyhow, I just wanted to see if this is right:
I found the final velocity to be (1.60m/s^2)(14.0s)= 22.4m/s
for the first 14 seconds. Then taking v-average to be 11.2m/s
and using d'x= 11.2 m/s * 14.0s= 156.8m for the first 14 seconds.
Then since it travels at a constant speed I'm assuming that v
average will still be 11.2m/s since it doesn't accelerate
during this period of 70 seconds. So I take d'x= 70s(11.2m/s)
and come to 784m for this period. Then when it begins to slow
down I will have 3.50m/s^2= 22.4m/s / d't, giving d't=
6.4seconds. Then take v average 11.2m/s= d'x / 6.4= 71.68m
Adding all these distances up you get 1012.48 meters.