Solve Mechanics Problems: Find Time & Displacement Relation

[AhmedHesham]

Homework Statement

A problem I can't solve

Relevant Equations

F=m(x+2)
F=m(2v+3)

First
F=m(x+2)
Initial velocity is 2m/s
find the relation between the time and displacement

Second
F=m(2v+3)
Initial velocity is 3 m/s
Find v(t)
Please help me
I tried but I can't finish it

 

[Chestermiller]

Problem Statement: A problem I can't solve
Relevant Equations: F=m(x+2)
F=m(2v+3)

First
F=m(x+2)
Initial velocity is 2m/s
find the relation between the time and displacement

Second
F=m(2v+3)
Initial velocity is 3 m/s
Find v(t)
Please help me
I tried but I can't finish it

What have you done so far?

 

[AhmedHesham]

What have you done so far?

first I applied a=vdv/dx
and in the second one
I applied a=dv/dt

 

[hilbert2]

Substitute an exponential trial solution

$A+Be^{Ct}$

(either as displacement or velocity) in the differential equation and try to determine what A,B,C should be for the equation to be correct.

 

[AhmedHesham]

Substitute an exponential trial solution

$A+Be^{Ct}$

(either as displacement or velocity) in the differential equation and try to determine what A,B,C should be for the equation to be correct.

How can I solve it with simple calculus

 

[berkeman]

How can I solve it with simple calculus

You need to start showing more effort here, or this thread will be closed. Please use the hint given by @hilbert2

 

[hilbert2]

The first equation is a differential equation for the displacement $x(t)$,

$m\frac{d^2 x(t)}{dt^2} = m(x(t)+2)$

and the second one a DE for the velocity $v(t)$ (it could also be written for displacement, but it's not what is asked here),

$m\frac{dv(t)}{dt} = m(2v(t)+3)$.

The initial condition in the first is $x(0)=2$, and in the second $v(0)=3$. The parameters in the equations are dimensionless, so I won't add units to these initial conditions, either.

If you put the exponential trial function in the place of $x(t)$ or $v(t)$ in either equation, you will get an equation that you can try to change into an identically true statement by adjusting $A,B,C$.

Sorry if I gave too much information compared to the poster's own effort.

 

[AhmedHesham]

The first equation is a differential equation for the displacement $x(t)$,

$m\frac{d^2 x(t)}{dt^2} = m(x(t)+2)$

and the second one a DE for the velocity $v(t)$ (it could also be written for displacement, but it's not what is asked here),

$m\frac{dv(t)}{dt} = m(2v(t)+3)$.

The initial condition in the first is $x(0)=2$, and in the second $v(0)=3$. The parameters in the equations are dimensionless, so I won't add units to these initial conditions, either.

If you put the exponential trial function in the place of $x(t)$ or $v(t)$ in either equation, you will get an equation that you can try to change into an identically true statement by adjusting $A,B,C$.

Sorry if I gave too much information compared to the poster's own effort.

thanks

 

[Chestermiller]

first I applied a=vdv/dx
and in the second one
I applied a=dv/dt

This was a good idea. So, in the first one, $$mv\frac{dv}{dx}=m(x+2)$$and$$vdv=(x+2)dx$$
Do you know how to integrate this?

 

[AhmedHesham]

This was a good idea. So, in the first one, $$mv\frac{dv}{dx}=m(x+2)$$and$$vdv=(x+2)dx$$
Do you know how to integrate this?

Yes
And I will have a relation between vsquard and x
What next

 

[Chestermiller]

Yes
And I will have a relation between vsquard and x
What next

Let’s see what you got.