05 Final Solutions

05 Final Solutions - 1. Velocity and Speed a. The figure...

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A 1. Velocity and Speed a. The figure shows a plot of the position of an object vs. time. Find the difference v – v x between object’s average speed ( v ) and average velocity ( v x ) during the time interval between t = 0.0 s and t = 7.0 s. v x = Δ x/ Δ t = (-5 m – 0 m)/(7 s) = -0.714 m/s v = d/ Δ t = (10m + 15 m)/(7 s) = 3.57 m/s v – v x = 4.29 m/s b. The figure shows a plot of the velocity of an object vs. time. If the object begins at position x = 0.00 m at time t = 0.00 s, find the time at which it passes through the point x = 30.00 m. (Note: there is no relation between this object’s motion and the graph in part a – do not use the graph from part a to solve this part of the problem!) Between t=0 and t=3 s, the object moves ½ (3 s) (8 m/s) = +12 m Between t=3 and t=5, the object moves (2 s) (8 m/s) = +16 m. Thus, at t=5, the object’s position is x = +28.0 m, and it has v x = +8.0 m/s After t=5.00 s the object’s acceleration is (-16.0 m/s)/(4.0 s) = -4.00 m/s 2 . It’s equation of motion after t=5.00 s is Δ x = ½ a ( Δ t) 2 + v 0 ( Δ t), where a = -4.00 m/s 2 , v 0 = 8.00 m/s, Δ t = t – 5.00 s and we want Δ x = 30.00m – 28.00 m = 2.00 m Solving the resulting quadratic gives Δ t = 0.268 s, so t = 5.27 s. The object also passes through x = 30.0 m at Δ t=3.73 s, t = 8.73 s (going in the other direction). Since the problem did not specify “the first time”, this answer is also correct.
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A 2. Acceleration and Free-Fall a. A block slides down a 0.500 m long frictionless plane inclined by 40.0° from horizontal, and then up another frictionless plane inclined at 30.0° from horizontal. How much time after release does it take for the block to come (momentarily) to rest? The block’s acceleration along the first plane ( θ 1 = 40°) is +g sin θ 1 . The acceleration along the second plane ( θ 2 = 30°) is –g sin θ 2 . Sliding down: Δ x 1 = 0.500 m = ½ (g sin θ 1 ) t 1 2 The block’s velocity at the bottom of the first plane v 1 = (2 g sin θ 1 Δ x 1 ) 1/2 Sliding up, we know the acceleration and can find the time for the block to lose all of its initial velocity: Δ v = -v 1 = a t 2 = -g sin θ 2 t 2 The total time from release is then t = t 1 + t 2 . Solving, we get t 1 = 0.398 s, v 1 = 2.510 m/s , t 2 = 0.512 s, t = 0.911 s b. A ball initially at rest is dropped from the top of a 30.0 m high building. At the same instant, another ball is thrown vertically upward from the ground with an initial speed of 25.0 m/s. Find the height at which the balls cross in mid air (ignore air resistance). Position of the first ball (dropped from rest): y 1 = h – ½ g t 2 , where h is the height of the building (= 30.0 m) Position of the second ball (thrown up from ground) y 2 = v 0 t – ½ g t 2 , where v 0 = 25.0 m/s Find the time of crossing: y 1 = y 2 , so h – ½ g t 2 = v 0 t – ½ g t 2 ; the quadratic terms cancel, hence h = v 0 t, and t = h/v 0 = 30.0 m/(25.0 m/s) = 1.20 s. The height of crossing can be found by plugging t into either equation for y:
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This note was uploaded on 09/08/2008 for the course PHYS 3A taught by Professor Casper during the Fall '07 term at UC Irvine.

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05 Final Solutions - 1. Velocity and Speed a. The figure...

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