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course Phy 241
1025pm 9/12/2012
RandProbAssign4Versn1#$&*
course Phy 241
415pm 9/6/2012
An automobile traveling a straight line is at point A at clock time t = 4 sec, where it is traveling at 10 m/s,
to point B at clock time t = 11 sec, where it is traveling at 20.5 m/s. Point A is 75 meters from the starting point
and point B is 115 meters from the starting point.
What are the average velocity and the average acceleration of the automobile during the specified time?
What evidence is there that the acceleration is or is not uniform?
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FROM POINT A TO B:
t_a = 4 seconds
v_a = 10 m/s
t_b = 11 seconds
v_b = 20.5 m/s
s_0 = 0 m
s_1 = 75 m
s_2 = 115 m
v_avg = (v_b + v_a) / 2
= (20.5 m/s + 10 m/s) / 2
= 15.25 m/s
a_avg = v_avg / 'dt
= 15.25 m/s) / 7 s
= 2.18 m/s^2
FROM STARTING POINT TO POINT B:
'dv = 'ds/'dt
= 115 m / 11 seconds
= 10.45 m/s
20.9 m/s - v_b = v_0 = .4 m/s
v_avg = (vb + v0) / 2
= [20.5 m/s + (20.5 m/s - 10.45 m/s)] / 2
= 15.275 m/s
a_avg = v_avg / 'dt
= (15.275 m/s) / 11 s
= 1.39 m/s^2
FROM STARTING POINT TO POINT A
v_avg = (va + v0) / 2
= (10 m/s + 10.05 m/s) / 2
= 10.025 m/s
a_avg = v_avg / 'dt
= (10.025 m/s) / 4 s
= 2.50625
The acceleration is not uniformed
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Assuming that the starting point is at 0 meters and clock time t = 0 s and the vehicle starts from rest.
t_start = 0 s
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It is not stated that anything starts at t = 0. The clock could have started at any time. The automobile could have been moving for years before the clock was started at some arbitrary time.
The point is that all you are given is information corresponding to two clock times. No other information is provided, so all information must follow from this given information with no extraneous assumptions.
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t_A = 4 s
t_B = 11 s
v_start = 0 m/s
v_A = 10 m/s
v_B = 20.5 m/s
x_start = 0 m
x_A = 75 m
x_B = 115 m
average velocity from starting point to point B
v_ave = 'dx/'dt
= (115m / 11s)
= 10.5 m/s
average velocity from starting point to point A
v_ave = 'dx/'dt
= (75 m / 4 s)
= 18.8 m/s
average velocity from point A to point B
v_ave = 'dx/'dt
= (115m - 75m)/(11s-4s)
= 14.5 m/s
average acceleration from starting point to point B
a_ave = 'dv/'dt
= (20.5 m/s / 11 s)
= 1.86 m/s^2
average acceleration from starting point to point A
a_ave = 'dv/'dt
= (10 m/s / 4 s)
= 2.5 m/s^2
average acceleration from point A to point B
a_ave = 'dv/'dt
= (20.5 m/s - 10 m/s) / (11 s - 4 s)
= (10.5 m/s / 7 s)
= 1.5 m/s^2
I was unclear what you meant by 'specified time', so I answer specified time from a number of different intervals.
The acceleration is NOT uniform based on the answers or average acceleration from starting point to point A,
which is 2.5 m/s^2 and average acceleration from Point A to Point B, which is 1.5 m/s^2
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I should have said 'during the specified time interval'.
Much, but not all of your analysis is based on the extraneous assumption that the car started with velocity 0 at clock time 0 and position 0. None of these assumptions can be made.
Can you revise your work to use only the given information?
Your information from point A to point B is all valid.
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Self-critique (if necessary):
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Self-critique rating:
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#$&*
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Once more you know nothing about the motion prior to point A, or after point B.
15.25 m/s is the average of the velocities at A and B. It is not the average velocity from A to B.
The average velocity is the average rate of change of position with respect to clock time, which leads to
vAve = `ds / `dt = (115 m - 75 m) / (11 sec - 4 sec) = 40 m / (7 sec) = 5.7 m/s, approx..
If acceleration was constant the v vs. t graph would be a straight line, the average velocity would occur at the midpoint of the interval, and would be equal to the average of initial and final velocities.
This isn't the case, so acceleration is not uniform.
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