# Position of an object after a certain amount of time?

• Loppyfoot
In summary, to find the position of the object 20 seconds after it passes the origin, you can use the kinematic equation s = ut + (1/2)at^2, where u is the initial velocity, a is the acceleration, and t is the time. In this problem, the initial velocity is <0,0,5> m/s, the acceleration is <0,0,-30> N/200kg, and the time is 20 seconds. By plugging in these values, you can find the final position of the object.
Loppyfoot

## Homework Statement

Your location is at the origin. When an object passes you, it is traveling 5 m/s in the −z direction. If the object's mass is 200 kg, and the net force on the object remains constant at ‹ 0, 0, −30 › N, what is the position of the object 20 seconds after it passes you?

## Homework Equations

The Change in momentum = Force * deltat

## The Attempt at a Solution

I solved for the impulse, and got the vector, <0,0,-600>N. This is equal to the final momentum minus the initial momentum. The initial momentum is <0,0,-8>m/s * 200kg. And the final momentum is pf = vf * 200kg. I solved for vf, using vf = vi + Fnet*deltat, but I am stuck.

I keep getting the wrong answer, and I am unsure of where I've went wrong and what I should do next. Thanks.

Loppyfoot said:

## Homework Statement

... what is the position of the object 20 seconds after it passes you?

...

## The Attempt at a Solution

I solved for the impulse, and got the vector, <0,0,-600>N. This is equal to the final momentum minus the initial momentum. The initial momentum is <0,0,-8>m/s * 200kg. And the final momentum is pf = vf * 200kg. I solved for vf, using vf = vi + Fnet*deltat, but I am stuck.

I keep getting the wrong answer, and I am unsure of where I've went wrong and what I should do next. Thanks.

You're looking for the final position, not the final velocity. Consider using a different kinematic equation, one that relates distance traveled to force applied to a mass (acceleration).

So I'll use:
deltax = vi*t + 0.5*a*t^2.
I plug in the proper values, and I don't get the correct answer. I am unsure of where I am going wrong. The initial position is, <0,0,0>, so I don't need to worry about that.

Ok,
xf = xi + vi*t + 0.5*a*t^2
xf = <0,0,0>m + <0,0,5>*20 + 0.5*(<0,0,-30>N/200kg)*(20s^2)

Check your velocity vector against what the original problem states.

So, the velocity would be <0,0,-5> since it is in the -z direction?

Loppyfoot said:
So, the velocity would be <0,0,-5> since it is in the -z direction?

What do you think?

you could have also used 2nd eqn of motion
s = ut + (1/2)at^2

## 1. What is the formula for calculating the position of an object after a certain amount of time?

The formula for calculating the position of an object after a certain amount of time is:
Position = Initial position + (Velocity x Time) + (1/2 x Acceleration x Time^2).

## 2. What units are used to measure position and time in this formula?

The units used to measure position are typically meters (m) or feet (ft), and the units used to measure time are usually seconds (s) or minutes (min).

## 3. Can this formula be used for any type of motion?

Yes, this formula can be used for any type of motion as long as the acceleration remains constant. If the acceleration changes, a different formula may need to be used.

## 4. What is the difference between position and displacement?

Position refers to the location of an object in relation to a fixed point, while displacement refers to the change in position of an object from its initial position to its final position.

## 5. How is the position of an object affected by changes in velocity or acceleration?

If the velocity changes, the position of an object will also change, as the object will be moving at a different rate. Similarly, if the acceleration changes, the position of the object will be affected as the object's speed and direction of motion will change.

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