A train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. Initially a car is 49 meters behind the train travelling in the same directionas the train at 37 m/s and accelerating at 4 m/s^2. What is the speed of the car just as it passes the train.

Sfg

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 setting up the following equation of motion for the car:

v = u + at

Where:
- v is the final velocity of the car (which we need to find)
- u is the initial velocity of the car (37 m/s)
- a is the acceleration of the car (4 m/s^2)
- t is the time it takes for the car to catch up to the train

Now, let's find the distance traveled by the car in terms of time. The car initially starts from 49 meters behind the train and has to cover this distance to catch up. The distance traveled by the car can be expressed as:

s = ut + (1/2)at^2

Where:
- s is the distance traveled (49 meters)
- u is the initial velocity of the car (37 m/s)
- a is the acceleration of the car (4 m/s^2)
- t is the time it takes for the car to catch up to the train

We can substitute the value of s in terms of t in the equation to get:

49 = (37)t + (1/2)(4)t^2

Simplifying the equation:

(1/2)(4)t^2 + (37)t - 49 = 0

Now, we can solve this equation for time t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = 37, and c = -49. Plugging these values into the equation:

t = [ -37 ± √(37^2 - 4(2)(-49)) ] / (2)(2)

t = [ -37 ± √(1369 + 392) ] / 4

t = [ -37 ± √1761 ] / 4

Calculating the square root:

t = [ -37 ± 41.96 ] / 4

Now, we need to consider the positive value of t since time cannot be negative for this problem. So, taking the positive value:

t = (4.96) / 4

t ≈ 1.24 seconds

Now that we have found the time it takes for the car to catch up to the train, we can calculate the final velocity of the car by substituting the value of t into the initial velocity equation:

v = u + at

v = 37 + (4)(1.24)

v ≈ 41.96 m/s

Therefore, the speed of the car just as it passes the train is approximately 41.96 m/s.