- 60 Views
- Uploaded on
- Presentation posted in: General

Distance-Rate-Time Applications

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Distance-Rate-Time Applications

Example 1:

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the rate on the ride home was 20mph faster than the rate going to work, what is the distance from her home to work?

1) Variable declaration:

Distance-Rate-Time Applications

Example 1:

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the rate on the ride home was 20mph faster than the rate going to work, what is the distance from her home to work?

1) Variable declaration:

Since the rate going home is in terms of the rate going to work, let x represent the rate going to work.

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the rate on the ride home was 20mph faster than the rate going to work, what is the distance from her home to work?

The rate returning home was 20mph faster than the rate going, or x+20.

The time going to work is 30 minutes, or …

The time returning home is 10 minutes, or …

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the ride home was 3 times the rate going to work, what is the distance from her home to work?

Since distance = rate × time, the distance to work is the product of the rate and time …

Do the same with the distance home …

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the ride home was 3 times the rate going to work, what is the distance from her home to work?

2) Write the equation

Since the distances are the same, we have …

3) Solve the equation:

4) Write an answer in words, explaining the meaning in light of the application

What was asked for in the application

Amy rides her bike to work in 30 minutes. On the way home she catches a ride with a friend and arrives home in 10 minutes. If the ride home was 3 times the rate going to work, what is the distance from her home to work?

x = rate riding to work

The rate riding to work was 10 mph.

1/2 hour = time riding to work

The distance from home to work was 5 miles.

Distance-Rate-Time Applications

Example 2:

Max leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Max is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Max’s rate.

1) Variable declaration:

Distance-Rate-Time Applications

Example 2:

Max leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Max is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Max’s rate.

1) Variable declaration:

Since Sam’s rate is given in terms of Mary’s rate, let x represent Mary’s rate.

Max leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Max is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Max’s rate.

Sam’s rate is 10 mph slower than Mary’s, or x-10.

Both Sam and Mary were traveling the same amount of time, from 11:00am to 3:00pm, which is 4 hours.

Since distance = rate × time, Sam’s distance is …

… and Mary’s distance is…

2) Write the equation

(Sam’s distance) + (Mary’s distance) = 480 miles

3) Solve the equation:

4) Write an answer in words, explaining the meaning in light of the application

What was asked for in the application

x =Mary’s rate

Mary’s rate was 65 mph.

Max’s rate was x – 10.

Max’s rate was 55 mph.

Distance-Rate-Time Applications

Example 3:

A plane travels against a 30mph wind for 3 hours. Then the plane travels with the same wind for 2 hours. The combined distance is 1270 miles. Determine the rate of the plane in still air.

1) Variable declaration:

Let x represent the rate of the plane in still air.

A plane travels against a 30mph wind for 3 hours. Then the plane travels with the same wind for 2 hours. The combined distance is 1270 miles. Determine the rate of the plane in still air.

When the plane is going against the wind, the ground speed is reduced by the rate of the wind.

The rate against the wind is given by …

(rate of the plane)

- (rate of the wind)

A plane travels against a 30mph wind for 3 hours. Then the plane travels with the same wind for 2 hours. The combined distance is 1270 miles. Determine the rate of the plane in still air.

When the plane is going with the wind, the ground speed is increased by the rate of the wind.

The rate with the wind is given by …

(rate of the plane)

+ (rate of the wind)

The time against the wind is 3 hours …

… and the time with the wind is 2 hours.

Since distance = rate × time, the distance against the wind is …

… and the time with the wind is…

2) Write the equation

(distance against the wind)+(distance with the wind) = 1270

3) Solve the equation:

4) Write an answer in words, explaining the meaning in light of the application

What was asked for in the application

x = rate of the plane

The plane’s rate in still air was 260 mph.

END OF PRESENTATION