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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 m/s. 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 roadkill on the runway are 93 meters and 132 meters at the respective clock times.Vave = (vf - v0) / (tf - t0) = ( 20.5 m/s - 10 m/s) / (15 - 8) = 5.57 m/s
Aave = 'dv /'dt = (20.5 m/s - 10 m/s) / (15 s - 8 s) = 1.5 m/s/s
The acceleration is uniform by this reasoning, which is hopefully the correct way to utilize the parallelogram law.
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Your calculation did not rely on uniformity of acceleration. You simply divided the change in velocity by the change in clock time, which by the definition of average acceleration gave you the average acceleration. Not other assumptions were necessary.
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In fact, if acceleration is constant, then the velocity vs. clock time graph is linear and the average velocity must be equal to the average of the initial and final velocities. However the average velocity is 5.57 m/s, while both the initial and final velocities are greater than this. Clearly the average velocity is not equal to the average of the initial and final velocities, and you must therefore conclude that the acceleration is not uniform.
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vave = 'ds / 'dt = (132 m - 93 m) / (15s -8s) = 5.57 m/s
This is my speed in the middle of the interval at t = 11.5 s
My area using my vave is to s = vave*'dt = 5.57 m/s * 7 s = 38.99 m = 39 m
So the slope between the initial point and final point represents the acceleration, and the area under the slope is the distance covered s.
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