Acceleration of a Truck Hit by a Water Stream

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SUMMARY

The discussion focuses on calculating the acceleration of a 2000kg truck hit by a horizontal water stream flowing at a rate of 600kg every 10 seconds with a velocity of 30 m/s. The thrust exerted by the water is determined using the equation thrust = (relative velocity of water to truck)(dM/dt), resulting in a non-constant acceleration due to the truck's increasing speed. The user initially encounters a conceptual error in their integration process, leading to an incorrect conclusion about the truck's speed. After clarification, they successfully derive the correct velocity equation and express intent to solve a subsequent part of the problem involving water retention in the truck.

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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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