Acceleration of a Truck Hit by a Water Stream

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In summary, the conversation discussed the problem of a hose producing a horizontal stream of water at a rate of 600kg every 10 seconds, striking the back of a 2000kg truck and causing it to accelerate. The thrust applied by the water was found to be dependent on the truck's speed and the acceleration was calculated using integration. A potential error in the integration was identified and corrected. The conversation also mentioned a part b of the problem, which involved the water entering the truck through a hole and remaining inside.
  • #1
pchalla90
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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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