Problem 1A-2: Average Acceleration
A tiny ice cube slides down a very slippery slope.
Along the slope at points A, B, C, D and E
devices are set up to measure the speed of
the ice cube and the time when the ice cube
passes the point. The results are tabulated below:
(a) What are the average accelerations between A and B, B and C, C and D, and D and E respectively?
(b) Can you estimate the distances between A and B, and between D and E?
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(a) Let the speed of the ice cube change by Δv m/sec during a time interval of Δt sec. The average acceleration during this time interval, ‹a›, is defined as ‹a› = Δv / Δt m/sec2. Thus
average acceleration between A and B = (4.00 - 0.00 m/sec) ÷ (1.00 - 0.00 sec) = 4.00 m/sec2. average acceleration between B and C = (12.00 - 4.00 m/sec) ÷ (3.00 -1.00 sec) = 4.00 m/sec2. average acceleration between C and D = (20.00 - 12.00 m/sec) ÷ (5.00 -3.00 sec) = 4.00 m/sec2. average acceleration between D and E = (32.00 - 20.00 m/sec) ÷ (8.00 -5.00 sec) = 4.00 m/sec2.(b) Between A and B:
The average speed, vAB = (4.00 + 0.00) ÷ 2 = 2.00 m/sec. The time needed to travel from A to B, tAB = 1.00 - 0.00 = 1.00 sec. The estimated distance between A and B is 2.00 × 1.00 = 2.00 m. Between D and E: The average speed vDE = (32.00 + 20.00) ÷ 2 = 26.00 m/sec. The time needed to travel from D to E, tDE = 8.00 - 5.00 = 3.00 sec. The estimated distance between A and B is 26.00 × 3.00 = 78.0 m. Note: If we assume that the acceleration through the whole movement is constant, that is, 4.00 m/sec2, it becomes possible to calculate exactly the distance between A and B, and between D and E respectively from the given data. We will learn later that the exact distance between A and B is 4.00 m, and between D and E is 156 m, about twice as large as our estimates. This is due to the presence of a large acceleration so that the speed within an interval is far from a constant but increases rapidly near the end of each interval.
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