Acceleration of a Truck Hit by a Water Stream

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Homework Help Overview

The problem involves a truck being accelerated by a horizontal stream of water from a hose. The truck has a mass of 2000 kg, and the water is delivered at a rate of 600 kg every 10 seconds with a constant speed of 30 m/s. The discussion focuses on determining the truck's acceleration over a period of 30 seconds, considering the effects of the water's momentum transfer.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the truck's acceleration using thrust equations and integration, questioning the physical feasibility of their results. Others engage in correcting integration errors and exploring the implications of different scenarios regarding water retention in the truck.

Discussion Status

Participants are actively discussing the mathematical approach to the problem, with some providing corrections to integration steps. There is a recognition of potential conceptual errors, particularly regarding the assumptions made about the system's behavior over time. The conversation is ongoing, with a focus on both parts of the question.

Contextual Notes

Participants are considering the implications of the water either splashing off the truck or being retained inside, which may affect the acceleration calculations. The original poster expresses uncertainty about their mathematical results, indicating a need for further clarification.

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Homework Statement


A hose can produce a horizontal stream of water at the rate of 600kg every 10 seconds as the water flows with constant speed of 30 m/s. Assume the water strikes the flat vertical back of a 2000kg truck. The water splashes out along the back of the truck without rebounding and then runs down to the ground. The truck is free to move without friction. What will be the acceleration of the truck in 30 seconds?


Homework Equations


thrust=|vdM/dt|
F=ma


The Attempt at a Solution



okay, so we know that dM/dt=600/10 kg/sec=60kg/sec.

we also know that as the truck speeds up, the relative speed of the stream of water will slow down, so the acceleration is not constant. this means that the thrust applied by the water is also not constant, but dependent on the v of the truck.

so here goes:

thrust=(relative v of water to truck)(dM/dt)=(30-v)(60)kgm/s^2

acceleration of truck=thrust/mass of truck=(1800-60v)/2000

a=dv/dt, so

dv/dt=(1800-60v)/2000
dt=2000dv/(1800-60v)

with a bit of integrating:

t=-(100/3)ln(90-3v)

and rearranging:

v=-(1/3)e^(-3t/100)-30

but when you plug in 30 seconds for t into this equation, you get a number greater than 30, which is physically impossible because the speed of the truck has v=30 as its asymptote.

what am i doing wrong?

thanks for your help.
 
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and another thing: assuming that i have a conceptual error and not a mathematical error, i will also need assistance solving part b of the question:

What will be the acceleration of the truck in 30 seconds if the water can enter the truck through a hole and remain inside?

thanks again.
 
Integrating your dt=2000dv/(1800-60v)
I get integral from v = 0 to v of 2000/60*dv/(30-v)
Let u = 30 - v, du = -dv to get
integral from u = 30 to 30-v of du/u
t = -2000/60[ln(30-v) - ln(30)]
60t/2000 = ln[30/(30-v)]
30/(30-v) = e^(-60t/2000)
v = 30 - 30*e^(-60t/m)
Note that at time zero, v = 0. In your answer you get v = -1/3 - 30 at time zero.
 
ahhhh! there we go. so errors with integration...

thank you for correcting me! it works now.
now off to figure out part b.

thanks again!
 

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