A train is moving parallel and adjacent to a
highway with a constant speed of 27 m/s. Initially a car is 41 m behind the train, traveling
in the same direction as the train at 39 m/s
and accelerating at 4 m/s
2
.
What is the speed of the car just as it passes
the train?
Answer in units of m/s
234
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 cover the distance between them.
Let's start by finding the distance traveled by the car during this time. We'll use the equation:
d = v₀t + 0.5at²
Where:
- d is the distance traveled
- v₀ is the initial velocity
- t is the time
- a is the acceleration
In this case, the initial distance between the car and the train is 41 m, the initial velocity of the car is 39 m/s, and the acceleration is 4 m/s². We want to find the time it takes for the car to cover this distance, so we can rearrange the equation:
t = (d - v₀t) / (0.5a)
Substituting the known values:
t = (41 - 39t) / (0.5 * 4)
Simplifying the equation:
2t = (41 - 39t) / 4
8t = 41 - 39t
47t = 41
t = 41 / 47
Now we have the time it takes for the car to pass the train. To find the speed of the car at this moment, we'll use the equation:
v = v₀ + at
Substituting the known values:
v = 39 + 4 * (41 / 47)
Simplifying the equation:
v = 39 + (164 / 47)
v = 39 + 3.49
Therefore, the speed of the car just as it passes the train is approximately 42.49 m/s.