High School

1. **Speeding Car and Police Pursuit Problem:**

A speeding car traveling north at 92 mph passes a road intersection where it is spotted by the police. After spending 34 seconds reporting, the police pursue the car at 100 mph. How many miles north will they be when the police catch up with the car?

2. **Trains and Superhero Problem:**

Two trains are approaching each other head-on on a straight track. Train A is on the left traveling right at 30 m/s. Train B is on the right traveling left at 10 m/s. When the trains are exactly 50 km apart, a superhero flies from the front of Train A to Train B.

a. If the superhero flies at 55 m/s, how far must they travel to go from Train A to Train B?

b. When the superhero reaches Train B, they turn and fly back to Train A at 55 m/s. How far must they fly to return?

c. When the superhero reaches Train A again, they fly back to Train B at 55 m/s. How far must they fly for the second trip?

d. If the superhero continues flying back and forth between the trains, how far will they have flown after making 3 round trips?

e. How many round trips will the superhero make before the trains collide, and what will be the total distance flown?

3. **Lifeguard Rescue Problem:**

A lifeguard is 100 meters down the beach from a drowning person 17 meters from shore. The lifeguard is standing 25 meters from the water's edge. His top running speed on sand is 8 m/s, and his swimming speed is 4 m/s.

a. If the lifeguard runs diagonally to a point on the shore perpendicular to the victim, then swims 17 meters, how long will it take to reach the victim?

b. If the lifeguard runs 17 meters directly to the water and then swims diagonally, how long will this path take?

c. Find the two diagonals that give the shortest time if the lifeguard runs along one diagonal on the shore and swims along another. Calculate the shortest total time.

Answer :

Final answer:

The police will be 0.94 miles north of the intersection when they catch up with the speeding car. The hero must fly 137,500 meters from Train A to Train B and back. The lifeguard will take 17.13 seconds to make contact by running diagonally and then swimming 17 meters.

Explanation:

(a) To find the distance that the police will travel to catch up with the speeding car, we need to calculate the time it takes for the police to catch up. The police are traveling at 100 mph, so in 34 seconds, they will cover a distance of:

distance = (speed) x (time) = 100 mph x (34/3600) hours = 0.94 miles

Therefore, the police will be 0.94 miles north of the intersection when they catch up with the speeding car.

(b) To calculate the distance the hero must travel from Train A to Train B, we can use the formula:

distance = speed x time = 55 m/s x (50,000 m / (30 m/s + 10 m/s)) = 137,500 m

(c) The hero must fly the same distance as in part (b), which is 137,500 m.

(d) After making 3 round trips, the hero will have flown a total distance of:

total distance = 2 x distance per round trip x number of round trips = 2 x 137,500 m x 3 = 825,000 m

(e) To determine the number of round trip flights the hero will have made by the time the trains collide, we can divide the initial distance between the trains by the distance per round trip:

number of round trips = (50,000 m) / (137,500 m) = 0.36 round trips

Since the hero cannot make a fraction of a round trip, the hero will have made 0 round trips by the time the trains collide. The total distance flown in all the trips will therefore be:

total distance flown = 2 x distance per round trip x number of round trips = 2 x 137,500 m x 0 = 0 m

(a) The lifeguard can use the Pythagorean theorem to calculate the time it takes to make contact. The distance the lifeguard needs to travel can be found using the theorem:

distance = sqrt((25 m)^2 + (100 m)^2) = sqrt(625 m^2 + 10,000 m^2) = sqrt(10,625 m^2) = 103.06 m

The total time to make contact is the time it takes to run across the beach plus the time it takes to swim 17 m:

total time = (distance / running speed) + (17 m / swimming speed) = (103.06 m / 8 m/s) + (17 m / 4 m/s) = 12.88 s + 4.25 s = 17.13 s

(b) In this case, the lifeguard only needs to swim diagonally to the victim. Using the Pythagorean theorem again, the distance is:

distance = sqrt((17 m)^2 + (100 m)^2) = sqrt(289 m^2 + 10,000 m^2) = sqrt(10,289 m^2) = 101.44 m

The time it takes to swim this distance is:

swimming time = distance / swimming speed = 101.44 m / 4 m/s = 25.36 s

(c) The shortest total time will be achieved by finding the diagonals of the rectangle formed by the lifeguard's initial position, the victim's position, and the perpendicular point on the shore. The two diagonals of this rectangle are:

diagonal 1 = sqrt((25 m)^2 + (100 m)^2) = sqrt(625 m^2 + 10,000 m^2) = sqrt(10,625 m^2) = 103.06 m

diagonal 2 = sqrt((17 m)^2 + (100 m)^2) = sqrt(289 m^2 + 10,000 m^2) = sqrt(10,289 m^2) = 101.44 m

The total time for this path is:

total time = (diagonal 1 / running speed) + (diagonal 2 / swimming speed) = (103.06 m / 8 m/s) + (101.44 m / 4 m/s) = 12.88 s + 25.36 s = 38.24 s

Learn more about Distance and Time here:

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