course Phy 201
I am unsure how to tell if the acceleration is uniform or if I am to use the following for vAve as opposed to below vAve (also denoted line over v):vAve=10m/s+20.5m/s/2=15.25 m/s?
vAve = `ds / `dt
If acceleration is uniform, then vAve is also equal to (vf + v0) / 2.
If vAve is not equal to (vf + v0) / 2, then acceleration is not uniform.
Be careful of order of operations.
If you write 10m/s+20.5m/s/2 then only the 20.5 m/s would be divided by 2; the division would occur before the multiplication.
What you mean is
(10m/s+20.5m/s) / 2,
which is equal to 15.25 m/s.
Clarification please
Assignment 4 Random problem
An airplane on a runway has velocity 10 m/s at clock time t-8 sec and velocity 20.5 m/s at clock time t = 15 s.
In the version of the problem you chose, t was given as 15 m/s.
However m/s is not a correct unit for velocity. The problem statement was in error. I've corrected the statement above.
It looks like this error led to your confusion. I apologize. See my notes above and below.
Find its average velocity and its average acceleration during this time interval and determine whether the acceleration might be uniform, if the distances from a certain road kill on the runway are 93 m and 132 m at the respective clock times.
v=10 m/s
t=8s
v=20.5 m/s
t=15 m/s
ā=v2-v1/t2-t1=change v/change t
ā=20.5m/s-10m/s/15m/s-8m/s=10.5/7=1.5 m/s
vAve=x2-x1/t2-t1=change x/change t
= 132m-93m/15m/s-8m/s=39/7=5.57 m/s
t is the clock time. 15 m/s would not be a clock time (should have read 15 s) and 8 s, not 8 m/s, is the other clock time. The units of clock time are seconds
Your first calculation should have read
(20.5m/s-10m/s) / (15s-8s) =10.5 m/s / (7 s) = 1.5 m/s^2.
This is the acceleration.
Your second calculation should have read
( 132m-93m )/ (15s-8s) = 39 m / (7 s) = 5.57 m/s.
This is the average velocity.
Note that I have specified units at every step, and the units calculations are correct.
What therefore is your conclusion about the uniformity of acceleration?