# Mechanics Problem: Finding Displacement, Velocity, and Acceleration Formulas

• lavalamp
In summary: However, this time, you should integrate the term v dv/dt with respect to t, which will require a u-substitution, where u is the function v(t).In summary, the conversation discusses a mechanics problem involving an idealized car with specific characteristics and the goal of finding formulae for displacement, velocity, and acceleration in terms of time. The attempt at a solution involves solving a second order non-linear differential equation, but the integration is done incorrectly. Suggestions are given for simplifying the problem or using numerical methods. The conversation also touches on the difficulty of solving non-linear odes and the possibility of finding solutions in engineering books. Finally, a summary of the incorrect integration and a suggestion to use u-substitution to correctly solve the
lavalamp

## Homework Statement

Hey all, this isn't actually a homework question, but I guess it's of that type. For some time now I've had this (not entirely realistic) mechanics problem that I keep leaving for a while, and then coming back to. Basically, I'm not getting anywhere so I am asking for some help with it. Ideally I'd like to find formulae for s, v and a in terms of time, and I'd appreciate any help people can offer.

It's an idealised car with power 100 kW, mass 1000 kg, and a seemingly infinite amount of grip. There is also an aerodynamic drag force which I've set at -0.5v^2. The car begins accelerating from rest at t=0 and I'd like to find it's displacement, velocity and acceleration at a given time.

## Homework Equations

I can get a formula for acceleration in terms of velocity relatively easily:

P = Tv
T = P/v

F = ma
T + Fd = ma
P/v - 0.5v2 = ma
100000/v - 0.5v2 = 1000a
200000/v - v2 = 2000a

a = (200000/v - v2)/2000

$$a = \frac{100}{v} - \frac{v^{2}}{2000}$$

The question is, what comes next?

## The Attempt at a Solution

Here are two equations I arrive at when I attempt to progress a little further, I'm fairly confident that they are both incorrect:

$$v^{3} = \frac{600000s}{3s + 2000}$$

$$s^{3}\ +\ 3000s^{2}\ +\ hs\ =\ 200000t^{3}\ +\ 6000000*5^{1/3}t^{2}\ +\ 20*5^{2/3}ht$$

c, e, f and h are constants. I actually couldn't eliminate h.

Here's how I came up with the first equation:

a = 100/v - v2/2000
2000a = 200000/v - v2
2000va = 200000 - v3

$$2000v\ \frac{dv}{ds}\frac{ds}{dt} = 200000 - v^{3}$$

$$2000v^{2}\ \frac{dv}{ds} = 200000 - v^{3}$$

$$\int 2000v^{2}\ dv = \int 200000 - v^{3}\ ds$$

$$\frac{2000v^{3}}{3} = 200000s - v^{3}s + c$$

$$v^{3}s + \frac{2000v^{3}}{3} = 200000s + c$$

3v3s + 2000v3 = 600000s + c

v3(3s + 2000) = 600000s + c

$$v^{3} = \frac{600000s + c}{3s + 2000}$$

When t=0, s=0 and v=0, therefore c=0.

$$v^{3} = \frac{600000s}{3s + 2000}$$

And the second one:

Calculating the maximum velocity of the car (used later):

P = Tv
P = 0.5v2 * v
P = 0.5v3
200000 = v3
v = $$\sqrt[3]{200000}$$ ~= 58.48 m/s

a = 100/v - v2/2000
2000a = 200000/v - v2
2000va = 200000 - v3

2000v dv/dt = 200000 - v3

$$\int 2000v\ dv = \int 200000 - v^{3}\ dt$$

1000v2 = 200000t + c - d3s/dt2

$$\int \int 1000v^{2}\ d^{2}t = \int \int 200000t + c \ d^{2}t - \int \int \int d^{3}s$$

$$\int \int 1000\ \frac{d^{2}s}{dt^{2}}\ d^{2}t = 100000t^{3}/3 + ct^{2} + et + f - s^{3}/6$$

$$\int \int 1000\ d^{2}s = 100000t^{3}/3 + ct^{2} + et + f - s^{3}/6$$

500s2 + hs = 100000t3/3 + ct2 + et + f - s3/6

s3/6 + 500s2 + hs = 100000t3/3 + ct2 + et + f

s3 + 3000s2 + hs = 200000t3 + ct2 + et + f

When t=0, s=0, therefore it's easy to spot right away that f=0.

When t is very large, v goes to cuberoot(200000) (max velocity calculated earlier), therefore s goes to cuberoot(200000)t. So setting s=cuberoot(200000)t

200000t3 + 3000*2000002/3t2 + 2000001/3th = 200000t3 + ct2 + et

6000000*51/3t2 + 20*52/3th = ct2 + et

c = 6000000*51/3
e = 20*52/3h

s3 + 3000s2 + hs = 200000t3 + 6000000*51/3t2 + 20*52/3ht

Hi,

To get your first equation, the integral is done wrongly. v is treated as a constant in the integral when it is not.

The equation you are trying to solve is a second order non-linear differential equation. a is a second order differential of s and v is a first order differential of s.

Thanks for the reply. Yeah, for the integration in the first one I had a bad feeling about that as I was doing it, but I didn't, and don't, know what else to do there.

Do you have any suggestions about how I can um ... not do it wrong? I'm pretty much at the limit of what I know here, which is not very much.

For the second equation I spotted a mistake, so starting from here again:

$$\int \int 1000v^{2}\ d^{2}t = \int \int 200000t + c \ d^{2}t - \int \int \int d^{3}s$$

$$\int \int 1000\ \frac{d^{2}s}{dt^{2}}\ d^{2}t = 100000t^{3}/3 + ct^{2} + et + f - \int \int s\ +\ g\ d^{2}s$$

$$\int \int 1000\ d^{2}s + \int \int s\ +\ g\ d^{2}s = 100000t^{3}/3 + ct^{2} + et + f$$

$$\int \int s + 1000 + g\ d^{2}s = 100000t^{3}/3 + ct^{2} + et + f$$

s3/6 + 500s2 + gs2 + hs = 100000t3/3 + ct2 + et + f

s3 + 3000s2 + gs3 + hs = 200000t3 + ct2 + et + f

s3 + (3000+g)s2 + hs = 200000t3 + ct2 + et + f

And again, when t=0, s=0, therefore f=0.

s3 + (3000+g)s2 + hs = 200000t3 + ct2 + et

Replacing constants:
a = 3000 + g
b = h

s3 + as2 + bs = 200000t3 + ct2 + et

And the trick I pulled last time, setting s equal to the cuberoot of 200,000 multiplied by t, I don't think is valid. Which means I'm stuck with 4 constants I don't know how to find the values of.

However, I ran four simple simulations of this car and found 4 values for s and t, from which I used simultaneous equations to find:

a = 2229.30
b = -252.85
c = -3673.88
e = -454.72

I'm not sure what the accuracy of these values is, I'd have to guess not too great since I ran a fifth simulation and solved the simultaneous equations again to get somewhat different values:

a = 2229.10
b = -257.97
c = -3731.14
e = -464.78

Well... non-linear odes are hard to solve, I'm not proficient at it either, and scientists often use numerical methods to calculate whatever they need if they can't solve the ode. Maybe you could simplify your problem to a power one dependence of the velocity on the drag force, or you could look up some engineering books to see how they did it.

If you require a numerical solution, look up the Runge-Kutta method.

The first integral is done wrongly for your second equation. integrating v^3 with respect to t is not d3s/dt3. I see that you are attempting to solve the same integral as with the first equation.

## 1. What is displacement and how do you calculate it?

Displacement is a measure of an object's change in position, usually measured in meters (m). To calculate displacement, you need to know an object's initial position (x0) and its final position (x). The formula for displacement is:
Δx = x - x0

## 2. How do you find an object's average velocity?

Average velocity is the rate of change of an object's position over a period of time. To find average velocity, you need to know the displacement (Δx) and the time interval (Δt). The formula for average velocity is:
vavg = Δx / Δt

## 3. What is acceleration and how is it related to velocity?

Acceleration is the rate of change of an object's velocity over a period of time. It is usually measured in meters per second squared (m/s2). Acceleration is related to velocity because it describes the change in an object's velocity. The formula for acceleration is:
a = Δv / Δt

## 4. How do you calculate an object's final velocity?

An object's final velocity is the velocity it has at a specific point in time. To calculate final velocity, you need to know the initial velocity (v0), the acceleration (a), and the time interval (t). The formula for final velocity is:
v = v0 + at

## 5. How can you use displacement, velocity, and acceleration formulas to solve real-world problems?

These formulas can be used to solve a variety of real-world problems, such as finding the distance traveled by a car, determining the speed of a falling object, or calculating the acceleration of a rocket. By plugging in known values and using the appropriate formulas, you can solve these problems and understand the motion of objects in the world around us.

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