What have you done so far?AhmedHesham said: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
first I applied a=vdv/dxChestermiller said:What have you done so far?
How can I solve it with simple calculushilbert2 said: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.
thankshilbert2 said: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.
This was a good idea. So, in the first one, $$mv\frac{dv}{dx}=m(x+2)$$and$$vdv=(x+2)dx$$AhmedHesham said:first I applied a=vdv/dx
and in the second one
I applied a=dv/dt
YesChestermiller said: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?
Let’s see what you got.AhmedHesham said:Yes
And I will have a relation between vsquard and x
What next
In order to find the time and displacement relation in a mechanics problem, you must first identify the relevant physical quantities, such as initial velocity, acceleration, and displacement. Then, you can use equations of motion, such as the kinematic equations, to solve for the time and displacement relation.
The kinematic equations are a set of equations that describe the motion of an object in terms of its initial velocity, acceleration, displacement, and time. These equations can be used to solve mechanics problems by plugging in known values and solving for the unknown variables.
Yes, the kinematic equations can be used for any type of motion as long as the acceleration is constant. This means that the velocity and acceleration cannot change over time.
If the acceleration is not constant, then the kinematic equations cannot be used to solve the problem. In this case, you may need to use other equations, such as the equations of motion for a variable acceleration, or use numerical methods to approximate the solution.
To check if your solution is correct, you can use the kinematic equations to plug in the values you have calculated and see if they match the initial values given in the problem. You can also check if your solution makes sense intuitively by considering the physical principles involved in the problem, such as conservation of energy or conservation of momentum.