A train is moving parallel and adjacent to a
highway with a constant speed of 22 m/s. Initially a car is 69 m behind the train, traveling
in the same direction as the train at 35 m/s
and accelerating at 3 m/s
2
.
What is the speed of the car just as it passes
the train
D1 = D2 - 69.
V*t = Vo + 0.5at^2 -69,
22t = 35t + 0.5 * 3t^2 - 69,
22t = 35t + 1.5t^2 - 69,
22t - 1.5t^2 - 35t = -69,
-1.5t^2 -13t + 69 = 0,
Multiply both sides by -1:
1.5t^2 + 13t - 69 = 0.
Using Quadratic Formula,
t = (-13 +- sqrt(169 + 414)) / 3,
t = (-13 +- 24.1) / 3,
t = 3.72, and -12.37.
Use the positive value of t:
t = 3.72s.
V = Vo + at = 35 + 3 * 3.72 = 46.12m/s.
To find the speed of the car just as it passes the train, we need to determine the position of the car at that moment.
Let's break down the problem and solve it step by step:
Step 1: Determine the time it takes for the car to catch up to the train:
To find the time it takes for the car to pass the train, we need to calculate the distance between them and divide it by the relative speed between them.
Relative speed = Speed of the car - Speed of the train
Relative speed = 35 m/s - 22 m/s
Relative speed = 13 m/s
Distance = Initial distance between the car and the train = 69 m
Time = Distance / Relative speed
Time = 69 m / 13 m/s
Time = 5.31 seconds (rounded to two decimal places)
Step 2: Find the final velocity of the car at that time:
The final velocity of the car can be found using the equation of motion.
Final velocity (vf) = Initial velocity (vi) + (acceleration * time)
Initial velocity (vi) = Speed of the car = 35 m/s
Acceleration (a) = 3 m/s^2
Time (t) = 5.31 seconds
vf = 35 m/s + (3 m/s^2 * 5.31 s)
vf = 35 m/s + (15.93 m/s)
vf = 50.93 m/s (rounded to two decimal places)
So, the speed of the car just as it passes the train is approximately 50.93 m/s.