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Get Full Access to Fundamentals Of Physics - 10 Edition - Chapter 9 - Problem 113
Get Full Access to Fundamentals Of Physics - 10 Edition - Chapter 9 - Problem 113

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# A railroad car moves under a grain elevator at a constant ISBN: 9781118230718 79

## Solution for problem 113 Chapter 9

Fundamentals of Physics | 10th Edition

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Problem 113

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 540 kg/min. What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

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Displacement, velocity and acceleration Ex. a particle moves back and forth along a straight line. Its displacement at time t ( t is in seconds, s(t)) is in feet is s(t)=t^3 ­ 12t^2 +36t a. What is the particle’s velocity at time t v(t) = s’(t) =3t^2 ­ 24t + 36 ft/sec b. When is the particle at rest Solve v(t)= 3t^2 ­ 24t + 36 = 0 Divide by three t^2 ­ 8t + 12 = 0 factor (t ­ 2)(t + 6) = 0 t = 2 sec, 6 sec c. what is the initial velocity Plug in 0 for t v(0) = 36 ft/sec d. what is the displacement at t =2 and t =6 sec Plug in 2 and 6 to the original function s(2) = 32 ft s(6) = 0ft Line graph Graph e. When is the particle moving in the positive direction When v(t) > 0 3t^2 ­ 24t + 36 > 0 Using the graph, 0 6 Interval notation: (0,2)U(6, ∞ ) f. Find the total distance the particle travels in the first 10 seconds. We need to find the distance traveled from t = 0 to t=2; t = 2 to t =6; and t = 6 to t = 10 [ ] = absolute value If the first 2 seconds: [s(2) ­ s(0) ] = [32 ­ 0 ] = 32 ft From t = 2 to t =6 [ s(6) ­ s(2)] = [ 10 ­ 32 ] = 32ft From t = 6 to t = 10 [ s(10) ­ s(6) ] = [160 ­ 0 ] = 160 ft Total distance = 32 + 32 + 160 = 224 ft g. Find the acceleration at time t. a(t) = v’(t) = s”(t) = 6t ­ 24 ft/s^2 h. When is the acce

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##### ISBN: 9781118230718

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