# Finding velocity in terms of displacement

• B
• Saracen Rue

#### Saracen Rue

I know two methods as to how to do this, but one of them doesn't seem to work.
Say you're given x = 3t^2 + 6t - 4.
If you treat x as an unknown variable and movie it over to the other side, you can use completing the square to solve for t. This eventually gets you: t = 1/3*sqrt(3(x+7)) - 1
Then you differentiate the initial function to obtain the velocity, substitute the calculated time in and thus obtain velocity in terms of displacement. This method ultimately got me v = 2*sqrt(3x+21), which works and is completely fine.

However, when I try the alternative way it doesn't work. If we go back to the beginning and find both velocity and acceleration, we get x = 3t^2 + 6t - 4, v = 6t + 6 and a = 6. To obtain dv/dx, we need to multiple the dv/dt by dt/dx (with the dt on the numerator and denominator canceling out). We know dv/dt, as that's just another way of expressing the acceleration, which is 6. Now this is the part which I think is messing up; I contend that dt/dx would should just be the inverse of dx/dt, which we know to be 6t + 6. This should mean that dt/dx should just equal 1/(6t + 6). Thus, dv/dx = dv/dt * dt/dx = 6*(1/(6t + 6)) = 1/(t+1). To obtain velocity in terms of displacement, we then would need to differentiate dv/dx, which should leave us with v(x) = ln(|t + 1|) + c. At t = 0, we know velocity = 6. Therefore, 6 = ln(|1|) + c, c = 6. v(x) = ln(|t+1|) + 6.

However, this equation is clearly very different from the one obtained in the first method and doesn't work. Can someone please tell me how to fix the second method?

I know two methods as to how to do this, but one of them doesn't seem to work.
Say you're given x = 3t^2 + 6t - 4.
If you treat x as an unknown variable and movie it over to the other side, you can use completing the square to solve for t. This eventually gets you: t = 1/3*sqrt(3(x+7)) - 1
Then you differentiate the initial function to obtain the velocity, substitute the calculated time in and thus obtain velocity in terms of displacement. This method ultimately got me v = 2*sqrt(3x+21), which works and is completely fine.

However, when I try the alternative way it doesn't work. If we go back to the beginning and find both velocity and acceleration, we get x = 3t^2 + 6t - 4, v = 6t + 6 and a = 6. To obtain dv/dx, we need to multiple the dv/dt by dt/dx (with the dt on the numerator and denominator canceling out). We know dv/dt, as that's just another way of expressing the acceleration, which is 6. Now this is the part which I think is messing up; I contend that dt/dx would should just be the inverse of dx/dt, which we know to be 6t + 6. This should mean that dt/dx should just equal 1/(6t + 6). Thus, dv/dx = dv/dt * dt/dx = 6*(1/(6t + 6)) = 1/(t+1). To obtain velocity in terms of displacement, we then would need to differentiate dv/dx, which should leave us with v(x) = ln(|t + 1|) + c. At t = 0, we know velocity = 6. Therefore, 6 = ln(|1|) + c, c = 6. v(x) = ln(|t+1|) + 6.

However, this equation is clearly very different from the one obtained in the first method and doesn't work. Can someone please tell me how to fix the second method?

I think you are given an equation for x as a function of time t ... however if you plot x as function of t you have initial condition for the motion described.
at t=0 the position of the body is at x= -4 and at t=t1 and t2 ,two values the body is at x=0 i.e. the origin as you have a quadratic in t.

one of them is real...
moreover if you place the origin at -4 the acceleration is 6 units and and the initial velocity is also 6 units.
so you have full description of the motion.
you are doing some exercises to again describe the motion .pl.check if you get a different result whether your initial condition is changed !

It is possible that ##\frac{dt}{dx}## is the derivative of ##t(x)=\frac{1}{3}\sqrt{3(x+7)}-1## respect ##x##, multiply this quantity by ##6## and look what happen ...

To obtain velocity in terms of displacement, we then would need to differentiate dv/dx, which should leave us with v(x) = ln(|t + 1|) + c. At t = 0, we know velocity = 6. Therefore, 6 = ln(|1|) + c, c = 6. v(x) = ln(|t+1|) + 6.

i think here you are making an error...
you have dv/dx = a function of t and now you are integrating dv on the left and dx . function of t on the right so you must first convert function of t in terms of x and then do the integration ...and i hope you will get consistent behaviour of velocity as function of x.