A train is moving parallel and adjacent to a highway with a constant speed of 22 m/s. Ini- tially a car is 35 m behind the train, traveling in the same direction as the train at 36 m/s and accelerating at 4 m/s2.

What is the speed of the car just as it passes the train?
Answer in units of m/s

car:

v = 36 + 4 t
x = 35 + 36 t + 2 t^2

train
v = 22
x = 22 t

the time t and the position x are the same at passing

22 t = 35 + 36 t + 2 t^2
solve quadratic for t
go back and find v = 36+4t

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 with the train. We can use the equation:

vf = vi + at

Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time

The initial velocity of the car (vi) is 36 m/s, and the acceleration (a) is 4 m/s^2.

Now, let's calculate the time it takes for the car to catch up with the train.

First, convert the initial distance between the car and the train to meters:

35 m

Next, we can use the equation:

vf = vi + at

Rearranging the equation to solve for time, we have:

t = (vf - vi) / a

Substituting the given values into the equation:

t = (22 m/s - 36 m/s) / 4 m/s^2
t = -14 m/s / 4 m/s^2
t = -3.5 s

We got a negative time, which indicates that we need to use the absolute value of the result since time cannot be negative.

Thus, the time it takes for the car to catch up with the train is 3.5 seconds.

Now, to find the final speed of the car just as it passes the train, we can use the equation:

vf = vi + at

Substituting the values into the equation:

vf = 36 m/s + (4 m/s^2)(3.5 s)
vf = 36 m/s + 14 m/s
vf = 50 m/s

Therefore, the speed of the car just as it passes the train is 50 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 define the position of the car as x_car and the position of the train as x_train. The initial position of the car is x_car = 35 m, and the initial position of the train is x_train = 0 m.

We can use the equation of motion for the car to relate its position, initial velocity, acceleration, and time: x_car = 35 + v_car_initial * t + 0.5 * a_car * t^2

Similarly, for the train, x_train = v_train * t

At the moment the car passes the train, their positions are the same, so we can set x_car = x_train and solve for the time (t).

35 + v_car_initial * t + 0.5 * a_car * t^2 = v_train * t

Substituting the given values: v_car_initial = 36 m/s, v_train = 22 m/s, a_car = 4 m/s^2, we have:

35 + 36 * t + 0.5 * 4 * t^2 = 22 * t

Rearranging the equation and simplifying:

2 * t^2 + 14 * t - 35 = 0

Now we can solve this quadratic equation for t. Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 2, b = 14, and c = -35.

t = (-14 ± √(14^2 - 4 * 2 * -35)) / (2 * 2)

Simplifying further:

t = (-14 ± √(196 + 280)) / 4

t = (-14 ± √476) / 4

Taking the positive square root:

t = (-14 + √476) / 4

t = (-14 + 2√119) / 4

t = -3.5 + 0.5√119

Since time cannot be negative in this context, we take the positive value:

t ≈ 0.67 seconds

Now that we have the time it takes for the car to catch up to the train, we can find the speed of the car just as it passes the train.

Using the equation of motion for the car:

x_car = 35 + v_car_initial * t + 0.5 * a_car * t^2

x_car = 35 + 36 * 0.67 + 0.5 * 4 * 0.67^2

Simplifying:

x_car ≈ 35 + 24.12 + 0.89

x_car ≈ 60.01 m

The speed of the car just as it passes the train is equal to the speed of the train, which is 22 m/s. So, the speed of the car just as it passes the train is 22 m/s.