280 likes | 383 Vues
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 t he rate on the ride home was 20mph faster than the rate going to w ork, what is the distance from her home to work?.
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.
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 time going to work is 30 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 rate on the ride home was 20mph faster than the rate going to work, what is the distance from her home to work? 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 …
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.
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. Both Sam and Mary were traveling the same amount of time, from 11:00am to 3:00pm, which is 4 hours.
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. Since distance = rate × time, Sam’s distance is … … and Mary’s distance is…
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. 2) Write the equation (Sam’s distance) + (Mary’s distance) = 480 miles
4) Write an answer in words, explaining the meaning in light of the application What was asked for in the application 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.
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)
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. The time against the wind is 3 hours … … and the time with the wind is 2 hours.
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. Since distance = rate × time, the distance against the wind is … … and the time with the wind is…
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. 2) Write the equation (distance against the wind)+(distance with the wind) = 1270
4) Write an answer in words, explaining the meaning in light of the application What was asked for in the application 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. x = rate of the plane The plane’s rate in still air was 260 mph.