Linear Motion Integration Problem

[VincentweZu]

Homework Statement

If the acceleration of a particle on a horizontal line is x where x represents the position in centimetres from the starting position of zero and if the initial velocity is 8 centimetres per second find the velocity of the particle when it is in a position of 15 centimetres

Homework Equations

The Attempt at a Solution

I attempted solving the equation using separation of variables.

let x = 15

a = 15
∫dv = ∫15dt
v = 15t + c c = 8
v = 15t + 8
∫dx = ∫15t + 8dt
x = 7.5t² + 8t Assuming object starts at origin c = 0
15 = 7.5t² + 8t Solving the quadratic
t = 0.9781

let t = 0.9781
v = 15(0.9781) + 8
v = 22.67cm/s

I know that something is wrong with my approach to the problem because this is not the answer, however I can't think of any other way to solve it. The answer is 17.

 

[SammyS]

Homework Statement

If the acceleration of a particle on a horizontal line is x where x represents the position in centimetres from the starting position of zero and if the initial velocity is 8 centimetres per second find the velocity of the particle when it is in a position of 15 centimetres

Homework Equations

The Attempt at a Solution

I attempted solving the equation using separation of variables.

let x = 15

a = 15
∫dv = ∫15dt
v = 15t + c c = 8
v = 15t + 8
∫dx = ∫15t + 8dt
x = 7.5t² + 8t Assuming object starts at origin c = 0
15 = 7.5t² + 8t Solving the quadratic
t = 0.9781

let t = 0.9781
v = 15(0.9781) + 8
v = 22.67cm/s

I know that something is wrong with my approach to the problem because this is not the answer, however I can't think of any other way to solve it. The answer is 17.

So, this is saying that the acceleration, a, is numerically equal to x, the position of the particle?

Of course, this does make an assumption about the units involved. You have position in units of cm. The more important unit is the unit of time, which is usually expressed in units of seconds.​

Yes it's true that when the particle is at position, 15 cm, the its acceleration is, 15 cm/s². However, you can't treat acceleration as if it's constant, which is what you did.

You need to solve the differential equation $\displaystyle\frac{d^2x}{d\,t^2}=x\, ,$ with the initial condition that the the velocity of the particle, dx/dt, is 8 cm/s, when t = 0 s.

 

[SammyS]

Assume that the velocity, v, is expressed as a function of position, x. This is natural, since we know acceleration as a function of position.

Using the chain rule with $\displaystyle a=\frac{dv}{dt}\,,$ gives $\displaystyle a=\frac{dv}{dx}\,\frac{dx}{dt}\,.$

Now replace dx/dt with v .

Can you solve the resulting differential equation?

 

[VincentweZu]

Thanks for your response!

So what I have to do is to start with

a = $\frac{dv}{dx}$$\frac{dx}{dt}$
a = $\frac{dv}{dx}$v

Since a = x

x = $\frac{dv}{dx}$v
∫(x)dx = ∫(v)dv
x²/2 = v²/2 + c where c = 8

Am I on the right track?

 

[namanjain]

(CGS unit system)
it is given
a=v dv/dx=x
so
v dv=x dx
integrating both sides

∫v dv=∫x dx where limit of vdv goes from '8 to v' and of xdx goes mrom '0 to x'

so
(v*2)/2 - (8*2)/2=(x*2)/2
on simplification it yields

(v*2) - 64 = (x*2)

now given x=15

v=√289=17cm/s




[I AM 99.99% SURE OF WHAT I've DONE]
[IT WAS DEFINATE INTEGRATION SO NO CONSTANT OF INTEGRATION]

 

[HallsofIvy]

Thanks for your response!

So what I have to do is to start with

a = $\frac{dv}{dx}$$\frac{dx}{dt}$
a = $\frac{dv}{dx}$v

Since a = x

x = $\frac{dv}{dx}$v
∫(x)dx = ∫(v)dv
x²/2 = v²/2 + c where c = 8

Am I on the right track?

You were until you said "c= 8"! Initially, v= 8 and x= 0. Putting those into your equation, 0²/2= 8²/2+ c so 32+ c= 0. c= -32, not 8.

Now set x= 15 and solve for v.