A train is moving parallel and adjacent to a highway with a constant speed of 28 m/s. Initially a car is 29 m behind the train, traveling in the same direction as the train at 42 m/s and accelerating at 3 m/s^2.
What is the speed of the car just as it passes the train?
d2 = d1+29
Vo2*t + 0.5a*t^2 = Vo1*t + 29 m
42t + 1.5t^2 = 28t + 29
1.5t^2 + 42t - 28t = 29
1,5t^2 + 14t = 29
1.5t^2 + 14t - 29 = 0
Use Quadratic Formula and get:
t = 1.745 s. To pass the train.
V = Vo + a*t = 42 + 3*1.745 = 47.24 m/s.
To find the speed of the car just as it passes the train, we need to determine the time it takes for the car to catch up to the train.
Let's start by looking at the relative velocity between the car and the train. Since they are moving in the same direction, we can subtract the velocity of the train from the velocity of the car:
Relative velocity = Car velocity - Train velocity
Relative velocity = 42 m/s - 28 m/s
Relative velocity = 14 m/s
Now, we can calculate the time it takes for the car to catch up to the train by using the equation:
Relative velocity = Acceleration * Time
14 m/s = 3 m/s^2 * Time
Solving for Time:
Time = 14 m/s / 3 m/s^2
Time ≈ 4.67 s
So, it takes approximately 4.67 seconds for the car to catch up to the train.
Now, we can use this time to find the speed of the car just as it passes the train. We can use the equation of motion:
Final velocity = Initial velocity + (Acceleration * Time)
Initial velocity of the car = 42 m/s
Acceleration of the car = 3 m/s^2
Time = 4.67 s
Plugging these values into the equation:
Final velocity = 42 m/s + (3 m/s^2 * 4.67 s)
Final velocity ≈ 42 m/s + 14 m/s
Final velocity ≈ 56 m/s
Therefore, the speed of the car just as it passes the train is approximately 56 m/s.