A train is moving parallel and adjacent to a

highway with a constant speed of 24 m/s. Initially a car is 48 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?

it's probably easiest to find out how long it takes, then recall that v = at.

So, if we have the car start the race at t=0, x=0,

35t + 3/2 t^2 = 48+24t
t = 3.07
so, v = 3*3.07 = 9.21 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 and pass the train.

Let's break down the problem step by step:

Step 1: Calculate the acceleration of the car.
The car is accelerating at a rate of 3 m/s^2.

Step 2: Determine the time it takes for the car to catch up to the train.
In order to catch up to the train, the car needs to close the initial distance of 48 m.
We can use the equation of motion:
s = ut + (1/2)at^2,
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Rearranging the equation, we get:
48 = 35t + (1/2)(3)t^2.

Simplifying the equation, we have:
48 = 35t + 1.5t^2.

Step 3: Solve the quadratic equation.
By rearranging the equation and setting it equal to zero, we get:
1.5t^2 + 35t - 48 = 0.

To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a,
where a, b, and c are the coefficients of the quadratic equation.

In this case, a = 1.5, b = 35, and c = -48.

By substituting these values into the quadratic formula, we get:
t = (-35 ± sqrt(35^2 - 4 * 1.5 * -48)) / (2 * 1.5).

Simplifying further, we have:
t = (-35 ± sqrt(1225 + 288)) / 3.

t = (-35 ± sqrt(1513)) / 3.

Step 4: Determine the value of t.
By solving the quadratic equation, we get two possible values for t:
t ≈ 3.08 s (positive value),
t ≈ -15.75 s (negative value).

Since time cannot be negative in this context, we discard the negative value.

Therefore, the car takes approximately 3.08 seconds to catch up to and pass the train.

Step 5: Calculate the speed of the car just as it passes the train.
To calculate the speed of the car just as it passes the train, we can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity of the car, u, is 35 m/s. The acceleration, a, is 3 m/s^2. The time, t, is 3.08 seconds.

Substituting these values into the equation, we get:
v = 35 + (3)(3.08).

Simplifying further, we have:
v ≈ 35 + 9.24.

Therefore, the speed of the car just as it passes the train is approximately 44.24 m/s.