Drawing a Force-Time Curve for a Changing Force and Calculating Final Momentum

[gold123456th]

A 1.0kg body initially traveling in the positive x-direction at 10m/s is acted upon for 2.0s by a force in that same direction of 20N. It then experiences a force acting in the negative direction for 20s equal to 2.0N. Draw a force-time curve and determine the final momentum

Homework Equations

I drew the 20N curve and 2.0N curve in the same coordinate plan, but I don't know if I should connect them with a vertical line. If I do, then at t=2s, the force would be both 20N and 2.0N

 

[jegues]

You should be able to draw the curve in one line without having your pencil leave the paper. (A continuous line)

Draw the curve after the first 2s then keeping your pencil on the paper, draw the curve for the next 20s, so you should end up with a domain of [0,22].

 

[gold123456th]

But then a vertical line would be drawn. Can vertical lines appear on a force-time graph? Then t= 2.0s would have a lot of forces.

 

[jegues]

Hmmm... I don't think it makes much of a difference as to whether you draw the line or not.

The question states that there is a 20N force for exactly 2 seconds then IMMEDIATELY after there is a 2N force for 20s.

You could always draw your horizontal 20N line then have an exclusive point at 2s on both lines then continue on with your 2N line. Because the we don't really know what is happening exactly at 2 seconds.