Motion Applications

Motion problems will involve the relationship of distance to rate and time.

Understand that rate(time) = distance or the formula r · t = d.  These problems will have two different things in motion.  The things in motion will either be going in opposite directions or in the same direction.  If they are traveling in opposite directions, you will add their distances that you find to form an equation where the total miles between them will be given.  If they are traveling in the same direction, you will subtract their distances that you find to form an equation where the difference (miles) between them will be given.

Example 1:  If a plane is going north at 500 miles per hour and another plane is going south at 550 miles per hour how long will it take for the planes to be 2625 miles apart.

Plane1 is going 500 miles per hour, so the rate is 500.  The time (T) is the unknown in this problem.  Therefore Plane1's distance will be 500T.

Plane2 is going 550 miles per hour, so the rate is 550.  The time (T) is the unknown in this problem.  Therefore Plane2's distance will be 550T.

Since the planes are going in opposite directions, then add their distances and set that total equal to how many mile apart they will be in T time.

500T + 550T  = 2625

1050T = 2625

T = 2.5 hours

Example 2:  Two cars leave Lewisburg both headed toward Nashville.  One car is traveling 50 mph and the other car is traveling 60 mph, how long will it take for the cars to be 5 miles apart?

Car1 is going 50 miles per hour, so the rate is 50.  The time (T) is the unknown in this problem.  Therefore Car1's distance will be 50T.

Car2 is going 60 miles per hour, so the rate is 60.  The time (T) is the unknown in this problem.  Therefore Car2's distance will be 60T.

Since the cars are going in the same direction, then subtract their distances and set that difference equal to how many mile apart they will be in T time.

60T - 50T  = 5

10T = 5

T = 0.5 hours (or 30 minutes)

Another type of motion problem is one that relates two objects in motion to their time.  If we solve the distance formula for time we would have the formula.
                                           or  simply          


Before we can set up another motion problem, let’s “think” about wind and current and how they affect rate.  Winds and currents can add to or subtract from rates depending upon whether the wind or current is working with or against the rate.  For example, if a boat is going downstream, the current would aid the boat working to help the boat go faster and would therefore be added to the rate of the boat in still water.  Going upstream the current would hinder the boat’s speed and would therefore be subtracted from the rate of the boat in still water.

Example 2:  A plane flies 350 miles with the wind in the same time that it can fly 310 miles against the wind.  The plane has a still-air speed of 165 miles per hour.  Find the speed of the wind.  

First, always identify what you are looking for with a variable.  Since we are looking for the speed of the wind, let’s let the speed of the wind = w.  We need to come up with two expressions for the times of the plane both “with the wind” and “against the wind”.  Since  , we can describe the times “with the wind” and “against the wind” in terms of their distances and rates, using the variable “w” to designate the wind's affect on the plane’s rate.

Time “with the wind” = 

Time “against the wind” = 

Now, set these two expressions for time equal to one another since the problem states “in the same time”.

                           … cross multiply to solve

    
     350(165-w) = 310(165+w)

     57750 – 350w = 51150 + 310w

      6600 = 660w

      w = 10 miles/hour  … the speed of the wind is 10 miles per hour!

Important hint:  When describing the rate that is affected by winds or current, the “still rate” whether it is known or unknown always comes first in the expression and the wind or current is then added or subtracted from the “still” rate whether it is known or unknown.  Getting these rates backwards will result in an incorrect equation!  For example if the still rate (r) is unknown and the wind is 2 mph then the rate “with the wind” would be r + 2; if the still rate is 250 mph and the wind (w) is unknown, then the rate “against the wind” would be 250 – w.
 

General Algebra Tips

The views and opinions expressed in this page are strictly those of Mary Lou Baker.
The contents of this page have not been reviewed or approved by Columbia State Community College.

This page was edited on 19-Sep-2007