How Does Constant Power Affect Truck Acceleration and Position?

[nns91]

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*t^3/2

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

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.

 

[tiny-tim]

Hi nns91!

… 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*t^3/2

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

erm … i think you need a power nap …

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

then v³ = … ?

 

[thomate1]

Hello,

Thank you for your question. It seems that you are on the right track by trying to relate the position and velocity equations. Here is a possible approach to solve this problem:

1. Start with the equation for position x = (8P/9m)^(1/2) * t^(3/2). This is the position equation that was previously found.

2. Take the derivative of this equation with respect to time, which will give you the velocity equation v = (2P/m)^(1/2) * t^(1/2).

3. Now, we can use the derivative to find the acceleration equation a = (P/m)^(1/2) * t^(-1/2).

4. We know that power is defined as P = F * v, where F is the force applied and v is the velocity. In this case, the force is constant and equal to P, so we can rewrite the equation as P = P * v. Solving for v, we get v = P/P = 1.

5. Now, we can plug this value of v into the acceleration equation to get a = (P/m)^(1/2) * t^(-1/2) = (P/m)^(1/2) * (1/t^(1/2)) = P/m.

6. Finally, we can use the equation for acceleration a = F/m to solve for the force F = ma = P.

7. Now, we can use the equation for work W = F * x to relate the power P to the position x. Substituting the value of F, we get W = P * x = P * (8P/9m)^(1/2) * t^(3/2).

8. Finally, we can use the definition of power P = W/t to relate the power P to the time t. Solving for t, we get t = W/P.

9. Substituting this value of t into the position equation x = (8P/9m)^(1/2) * t^(3/2), we get x = (8P/9m)^(1/2) * (W/P)^(3/2).

10. Simplifying this equation, we get x = (m/3P) * W^(3/2) = (m/3P) * (P * t)^(