A train is moving parallel and adjacent to a highway with a constant speed of 27 m/s. Ini- tially a car is 29 m behind the train, traveling in the same direction as the train at 38 m/s and accelerating at 2 m/s2.
What is the speed of the car just as it passes the train?
Answer in units of m/s
train goes 27 t meters
car goes 27 t + 29 meters
so
27 t + 29 = 38 t + (1/2)(2) t^2
N identical resistors are connected in parallel. The resistance of one of them is equal to R. What is the equivalent resistance of all N resistors?
To find the speed of the car just as it passes the train, we need to determine at what time the car catches up to the train. This can be done by setting up a relative motion problem and solving for the time it takes for the car to catch up.
First, let's determine the relative speed between the car and the train. Since they are moving in the same direction, the relative speed is the difference between their speeds:
Relative speed = Speed of car - Speed of train
Relative speed = 38 m/s - 27 m/s
Relative speed = 11 m/s
Now we can use the relative speed and the initial distance between the car and the train to calculate the time it takes for the car to catch up:
Relative distance = Initial distance between car and train
Relative distance = 29 m
Time = Relative distance / Relative speed
Time = 29 m / 11 m/s
Time = 2.636 s
Now that we know the time it takes for the car to catch up to the train, let's calculate the speed of the car at that moment. Since the car is accelerating, we can use the following kinematic equation:
Final velocity = Initial velocity + (Acceleration * Time)
Initial velocity of the car = 38 m/s (given)
Acceleration of the car = 2 m/s^2 (given)
Time = 2.636 s (calculated)
Final velocity = 38 m/s + (2 m/s^2 * 2.636 s)
Final velocity = 38 m/s + (5.272 m/s)
Final velocity = 43.272 m/s
Therefore, the speed of the car just as it passes the train is approximately 43.272 m/s.