An elevated train on a track 20 feet above the ground crosses a street at a rate of 25 feet per second and at a 30 degree angle. Five seconds later a car crosses under the tracks going 40 feet per second. How fast are the train and the car separating 3 seconds later?
To solve this problem, we first need to find the distances traveled by the train and the car.
We can use the formula:
Distance = Rate × Time
1. Distance traveled by the train in 5 seconds:
Distance_train = Rate_train × Time_train
Distance_train = 25 feet/second × 5 seconds
Distance_train = 125 feet
2. Distance traveled by the car in 5 seconds:
Distance_car = Rate_car × Time_car
Distance_car = 40 feet/second × 5 seconds
Distance_car = 200 feet
Next, we need to determine the distances covered by the train and the car during the 3-second interval after the 5-second mark.
3. Distance traveled by the train in 3 seconds:
Distance_train_3s = Rate_train × Time_train_3s
Distance_train_3s = 25 feet/second × 3 seconds
Distance_train_3s = 75 feet
4. Distance traveled by the car in 3 seconds:
Distance_car_3s = Rate_car × Time_car_3s
Distance_car_3s = 40 feet/second × 3 seconds
Distance_car_3s = 120 feet
Now, we can determine the distance between the train and the car after 3 seconds.
5. Distance between the train and the car after 3 seconds:
Distance_separation = Distance_train + Distance_car - (Distance_train_3s + Distance_car_3s)
Distance_separation = 125 feet + 200 feet - (75 feet + 120 feet)
Distance_separation = 125 feet + 200 feet - 75 feet - 120 feet
Distance_separation = 130 feet
Therefore, the train and the car are separating at a rate of 130 feet after 3 seconds.