Motion up an Inclined Plane

## Question:

Hey! I'm having trouble answering this question from my physics class, and I'm hoping that maybe you can help me solve this problem.

The dots below represent the position of a projectile every 0.10 s as it is projected at an angle up an inclined air table.

How do I calculate the average velocity during time interval (m/s) for the horizontal, and vertical axes.

Thanks... Hoping to hear from you soon!

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In general to find the average velocity you divide the total distance traveled by the total time. Then to get the horizontal component of that distance multiply the result by the cosine of the angle that the inclined plane makes with the horizontal. The vertical component is the total average velocity times the sine of that angle.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.

JDJ

## Question:

I'm sending you the same graph in two different formats (*.bmp, *.gif) this time. I understand what you are saying below, but can I get an example or two! I just want to make sure that I'm doing it the right way.

The dots in the graph represent the position of a projectile every 0.10s as it is projected at an angle up an inclined air table.

How do I calculate the average velocity during time interval (m/s) for the horizontal, and vertical axes.

I appreciate your help... Thanks a lot! This looks like one of those questions where you seem to be missing some important information... the angle of the inclined plane. Here is what I would do. By the way always check my arithmetic and logic. I dash these things off as fast as I can type and frequently screw up.

Since we know the acceleration, g, due to gravity may be taken as constant at -9.8 m/s/s in the vertical direction, the component of g that lies along the inclined plane, gy, will be g*sin(q) where q is the angle the plane makes with the horizontal. Assuming that your data is in meters measured along the inclined plane, that gy component of g is responsible for the decrease in velocity along the plane.

As long as the plane's angle remains fixed, the gy must also be constant. The velocity of an object subject to constant acceleration changes linearly with time so that we may use the rate of change in velocity between any two points to determine the magnitude of gy. The velocity at time=0.2 seconds is just the average velocity between time=0.0 and time=0.4, again because the rate of change of velocity is linear. That value is (0.8-0.0)/(0.4-0.0)=2m/s. Between time=0.4 and time=0.8 the average velocity is (1.45-0.8)/(0.8-0.4)=0.65/0.4=1.625m/s... which is also the velocity at time 0.6 seconds.

So between 0.2 seconds and 0.6 seconds the velocity decreased by 0.375 m/s, yielding an acceleration gy of -0.375m/s / 0.4s. Or gy=0.9375. Plugging this value into gy=g*cos(q) we get sin(q)=gy/g=0.9375/9.8=0.09566. That gives us q=5.5 degrees roughly.

Now we can grab any time interval we want, say from t=0.0 to t=4.0, find the average velocity by V=(4.0-0.0)/(4.0-0.0)=1.0m/s and calculate the horizontal and vertical components of that average by Vh=V*cos(q) and Vv=V*sin(q).

Regards,

JDJ