# Need to find a constant in an x-axis equation

• mcdowellmg

## Homework Statement

A 2.2 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s)t + ct2 - (2.6 m/s3)t3, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

F = m*a

## The Attempt at a Solution

I was given the hint: Acceleration is the second time derivative of the position function. The net force is related to the acceleration via Newton's second law.

I tried to take the 2nd derivative of the x equation, but I am not sure if I am doing it right because of all of the m's and m/s's, etc. along with the c and the t's.
I ended up getting 4+c*2t+t^2-7.8t^2 and then 2c-11.6t as my second derivative. (I had 2c-15.6t at first, but I think I had the product rule wrong in the first derivative then...)
I then solved a = 2c - 11.6(3) as 16.364 = 2c - 34.8, because 16.364 is the acceleration I found by dividing 36/2.2 (F/m). That solves for c as 25.58200. Apparently that is wrong, and I don't know why!

Because the position has a $$t^3$$ term, the force must be time dependent. The $$t^2$$ term must arise from the constant part of the force, but you must use the $$t^3$$ term to isolate that part. The best way to do this is to derive this position function from Newton's law using a time-dependent force.

I think that I may have forgotten that the 36 N is negative...which would be -36/2.2...so -16.364 = 2c - 34.8 would give me a different c, which I believe may be correct.

## Homework Statement

A 2.2 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s)t + ct2 - (2.6 m/s3)t3, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

F = m*a

## The Attempt at a Solution

I was given the hint: Acceleration is the second time derivative of the position function. The net force is related to the acceleration via Newton's second law.

I tried to take the 2nd derivative of the x equation, but I am not sure if I am doing it right because of all of the m's and m/s's, etc. along with the c and the t's.
I ended up getting 4+c*2t+t^2-7.8t^2
I presume you mean you got this as the first derivative- the velocity. Is that "t^2" a typo? It shouldn't be there and its derivartive, 2t, does not appear in your second derivative.

and then 2c-11.6t as my second derivative. (I had 2c-15.6t at first, but I think I had the product rule wrong in the first derivative then...)
No, the derivative of x= 3.0 m + (4.0)t + ct^2 - (2.6)t^3 is dx/dt= 4+ 2ct- 7.8t^2 and the second derivative is d^2x/dt^2= 2\c- 15.6t. 2(7.8)= 15.8. Even if that "t^2" in the first derivative was intended and you included it with "-7.8t^2" to get "-6.8t^2" the dervative would be "-13.6t", not "-11.6t".

I then solved a = 2c - 11.6(3) as 16.364 = 2c - 34.8, because 16.364 is the acceleration I found by dividing 36/2.2 (F/m). That solves for c as 25.58200. Apparently that is wrong, and I don't know why!
You should have 2c- 15.6(3)= 2c- 46.8= (-36/2.2)= -16.36 to two decimal places.