What Is the Velocity After 1 Meter?

[mailhiot]

If an object starts with an initial velocity of 0, and has a velocity of 1.07 m/s after 20 cm, what is the velocity at 1 meter, assuming a constant acceleration? I do not know how long it takes to travel the full 1 meter. All I know is that after 20 cm the velocity was 1.07 m/s and it took 177 miliseconds to travel the 20 cm.

 

[victor.raum]

Firstly, recall the general relationship between a constant force and a change in kinetic energy: $$F\cdot\Delta x = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$

In this case you have $v_1=0$, so that comes out to simply $$F\cdot\Delta x = \frac{1}{2}mv_2^2$$

Now find out what $F$ is by rearranging this to

$$F = \frac{\frac{1}{2}mv_2^2}{\Delta x}$$

Plug in any available velocity and corresponding location into the above, and you'll find the constant force.

Next rearrange to instead solve for the velocity as a function of distance

$$\sqrt{\frac{2F\cdot\Delta x}{m}} = v_2$$

And then flip that, because it just looks better flipped:

$$v_2 = \sqrt{\frac{2F\cdot\Delta x}{m}}$$

Plug in a value of 1 meter for the distance, and you'll find your velocity at that location.

Your textbook didn't explain this?

 

[HallsofIvy]

Equivalently, if an object moves with constant acceleration a m/s^2, starting with 0 velocity, after t seconds it will have velocity at m/s and will have traveled (1/2)at^2 m.

It will have traveled .2 m when (1/2)at^2= .2 so t= sqrt(.4/a) seconds. If, at that time, it has velocity 1.07 m/s, we have at= a(sqrt(.4/a))= 1.07. Squaring both sides, a^2(.4/a)= .4a= 1.07^2. Solve that for a.

Once you know a, use (1/2)at^2= 1 to determine the time the object has traveled to 1 m, then put that time and a into at to find the velocity