Monday, January 16, 2017

Motion with Constant Velocity or Speed

Motion with Constant Velocity or Speed

Motion with constant velocity or speed is, mathematically, a basic rate problem. That is to say that something occurs  
regularly over a period of time, and there is an outcome that is in relationship with a rate of "going on" and how long something goes on.

For example, if a cookie machine can turn out 10 cookies every 15 minutes, then the total number of cookies will be determined by how long the cookie machine runs. The rate can be stated as 40 cookies per hour, so after 3 hours, there will be 120 cookies.

Motion with a constant velocity or speed works the same way. The rate will be how far something moves in a given time—which is called speed or velocity—and time will be how long it moves.

In most cases, calculating any part of the relationship is relatively simple. It is not that hard to figure out a formulaic relationship by just looking at the units. (See Video.)

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A speedometer shows speed to be some number of miles per hour or kilometers per hour. So:

speed = KPH

The "per" means divide, so…

speed = k/h

where k is kilometers and h is hours. Changing to the variables normally used, the distance equation is:

d = vt

d is displacement, v is average velocity, and t is elapsed time.

Similarly, the same equation works for motion NOT along a straight line but with these changes:

d is distance traveled, v is average speed, and t is still elapsed time.

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A car travels at an average speed of 35 miles per hour for two hours. How far does it travel?

Begin by identifying the needed values:

d = what is being looked for 
v = 35 miles per hour 
t = 2 hours

d = vt 
d = 35 m/hr * 2 hr

d = 70 miles (the 'hr's cancel out)


A car travels at between two points that are 10 miles apart in a straight line. The trip begins at 8:00 and ends at 10:00. What is average velocity?

Begin by identifying the needed values:

d = 10 miles 
v = what is being looked for

t = final time - initial time
t = 10:00 - 8:00
t = 2 hours

d = vt 
10 miles = v * 2 hr
10 miles/2 hours = 
5 m/hr = v

The math for motion with constant velocity or speed is very easy, but rearranging the formula for each case could benefit some people. Thus, each part of the formula can be found using these three equations:

d = vt
v = d/t
t = d/v

Based on what is given in the problem, the math involves only multiplying or dividing, in most cases.

Naturally, there has to be a more complicated version of motion with constant velocity or speed. Nevertheless, it is not all that much more complex. It just looks more complex!

The more complex version asks how far something is from somewhere if it starts some distance away from some other place. IKR?

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Here is what that looks like:

A canoe left the shore 10 miles downstream from a bridge. It traveled at 2 miles per hour for 3 hours. How far was it from the bridge?

Now, to reason this out with logic alone, it is fairly easy. The canoe travels at 2 miles per hour for 3 hours, so it goes 6 miles. However, it STARTED 10 miles from the bridge, so it ends up a TOTAL of 16 miles from the bridge. BAM!

But the "official physics" formula uses subscripts. So… there's that. Or this:

df = di + vt


df is final distance/displacement (this is sometimes dt for total distance)
di is initial distance/displacement
v is average speed/velocity
t is time

Back to the canoe problem above, we have:

df = what we want to find
di = 10 miles from the bridge = 10 miles
v = 2 miles per hour
t = 3 hours

df = di + vt
df = 10 miles + 2 m/hr * 3 hr
df = 10 miles + 6 miles
df = 16 miles

The subscripts just make it look harder, but it is really just the same logic and reason. The math is just as easy—addition and multiplication.

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It is not really hard to rearrange the more complicated formula to solve for any of the four values. This is what it would look like:

df = di + vt
d= df - vt
t = (df - di)/v
v = (df - di)/t

While there are, indeed, formulas that model motion with constant speed or velocity, it is most important to remember that ration and logic always apply. It is tempting, at times, to give up because the formula is intimidating, but there is no need to do that. Remember that the formulas are just a way to get to the same conclusions that can be found by simply reasoning it out.

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