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Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?
Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?
Answer:a) The final velocity is 28.0 m/sb) It will take 50.9 s to come to a stop from a velocity of 28.0 m/sc) In the first case ( point "a") the train will travel 7.68 × 10³ m. In the secid case, the train will travel 713 mExplanation:The equations for the position and velocity of objects moving in a straight line are as follows:x = x0 + v0 • t + 1/2 • a •t²v = v0 + a • tWherex = position at time tx0 = initial positionv0 = initial velocityt = timea = accelerationv = velocity at time ta) Using the equation of velocity:v = v0 + a • tv = 4.00 m/s + 0.0500 m/s² • 480 sv = 28.0 m/sb) Now, the initial velocity is 28.0 m/s and the final velocity is 0 m/s. Then:0 m/s = 28.0 m/s - 0.550 m/s²• t-28.0 m/s / - 0.550 m/s² = tt = 50.9 sc) Now, we have to use the equation for the position of the train.x = x0 + v0 • t + 1/2 • a •t² (let´s place the center of the frame of reference at x0. In this way, x0 = 0)In the first case:x = 4.00 m/s • 480 s + 1/2 • 0.0500 m/s² • (480 s)²x = 7.68 × 10³ mIn the second case:x = 28.0 m/s • 50.9 s - 1/2 • 0.550 m/s² • (50.9 s)²x = 713 m...