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CP CALC A 2540-kg test rocket is launched vertically from

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 22E Chapter 5

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 22E

CP CALC? A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by v(t) = At + Bt2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?

Step-by-Step Solution:

Solution 22E Step 1 of 6: (a) Determine A and B , including their SI units. Given vertical velocity as the function of time, v = At + Bt ……….1 Solving for A, Differentiating with respect to time, d d 2 dtv = dt(At + Bt ) Using a= v d a = d(At) + (Bt ) 2 dt dt dt a= A+ 2Bt 2 Substituting at t= 0s , a= 1.5m/s 1.5m/s = A+ 0 2 Therefore, A= 1.5m/s Step 2 of 6: Solving for B Substituting A= 1.5m/s ,v=2 m/s at t=1s in equation 1, 2 v = At + Bt 2 m/s=1.5m/s (1s)+ B(1s) 2 2 B (1s) = 2 m/s -1.5 m/s B (1s) = 0.5 m/s B =0.5 m/s 3 3 2 Using B =0.25 m/s and A= 1.5m/s in equation 1 v = (1.5m/s )t + (0.5 m/s )t ………...2 Similarly acceleration equation becomes, a= (1.5m/s )+ 2(0.5 m/s )t 3 a= 1.5+ t…………..3 Step 3 of 6: (b) t 4.00 s after fuel ignition, what is the acceleration of the rocket, In order to calculate the acceleration at t=4s, we need to calculate the velocity at the same instant. Substituting t = 4s in equation 2, v = (1.5m/s )(4 s) + (0.5 m/s )(4 s) 2 V = 8 m/s By fundamental relation of acceleration, a= v t Using V = 7 m/s and t= 4s 8 m/s a = 4 s 2 a= 2 m/s 2 Therefore, acceleration 4s after the fuel ignition is 2 m/s .

Step 4 of 6

Chapter 5, Problem 22E is Solved
Step 5 of 6

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 22E from chapter: 5 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: Fuel, thrust, rocket, its, upward. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 22E from 5 chapter was answered, more than 960 students have viewed the full step-by-step answer. The answer to “CP CALC? A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by v(t) = At + Bt2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?” is broken down into a number of easy to follow steps, and 145 words.

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