How Does Constant Power Affect Truck Acceleration and Position?

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The discussion focuses on deriving the relationship between the position and speed of a truck accelerating with constant power. The initial equations provided are x=(8P/9m)^(1/2)t^(3/2) for position and v=(2P/m)^(1/2)t^(1/2) for speed. The goal is to show that x can be expressed as x=(m/3P)v^3. Participants are attempting to connect these equations through integration, but one user expresses difficulty in making the necessary connections. The conversation highlights the challenge of relating speed and position under constant power conditions.
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SUPER URGENT ! Power question

Homework Statement



A truck of mass m is accelerated from rest at t=0 with constant power P along a level road

show that the position of the truck is related to its speed by x=(m/3P)v^3

the position of the truck is previously found as:

x=(8P/9m)1/2*t3/2

I also found the v=(2P/m)1/2*t1/2

Homework Equations



Integral

The Attempt at a Solution



I have not found anyway to do this problem yet. I tried to to relate v to x by taking the integral of v by integrate (m/3P)v^3 but I am stuck.
 
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Hi nns91! :smile:
nns91 said:
… show that the position of the truck is related to its speed by x=(m/3P)v^3

the position of the truck is previously found as:

x=(8P/9m)1/2*t3/2

I also found the v=(2P/m)1/2*t1/2

erm … i think you need a power nap :redface: …

if v=(2P/m)1/2*t1/2,

then v3 = … ? :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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