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?

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 assume that when the car passes the train, they both have traveled a distance 'd'.

Since the train has a constant speed of 24 m/s, it takes the train the same amount of time as the car to cover the distance 'd'.

To find the time it takes for the car to cover the distance 'd', we can use the equation:

d = initial velocity * time + 0.5 * acceleration * time^2

For the car:
initial velocity (u) = 35 m/s
acceleration (a) = 3 m/s^2

For the train:
initial velocity (u) = 24 m/s
acceleration (a) = 0 (since the train has a constant speed)

Substituting the values into the equation for the car:

d = 35t + 0.5 * 3 * t^2

Substituting the values into the equation for the train:

d = 24t

Since both equations represent the same distance 'd', we can equate them:

35t + 0.5 * 3 * t^2 = 24t

Simplifying the equation:

1.5t^2 + 11t = 0

Factoring out the common factor of t:

t(1.5t + 11) = 0

Applying the zero-product property, we have two possible solutions:

1) t = 0
2) 1.5t + 11 = 0

Since we are looking for a positive time value, we can discard t = 0.

Solving for t:

1.5t + 11 = 0
1.5t = -11
t = -11/1.5
t ≈ -7.33

Since having a negative time value is not physically meaningful, we can discard it as well.

Therefore, the car passes the train at a positive time value.

Substituting t = 7.33 into the equation for the train distance:

d = 24 * 7.33
d ≈ 175.92 m

Therefore, the speed of the car just as it passes the train is the distance 'd' divided by the time 't':

Speed = d/t = 175.92 / 7.33
Speed ≈ 24 m/s

Hence, the speed of the car just as it passes the train is approximately 24 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 cover the distance between itself and the train.

First, let's calculate the initial distance between the car and the train: 48 meters.

Next, let's find the time it takes for the car to cover this initial distance. We can use the equation of motion:

distance = initial velocity * time + (1/2) * acceleration * time^2

Considering the initial velocity of the car as 35 m/s, the acceleration as 3 m/s^2, and the distance to be covered as 48 meters, we can rearrange the equation to solve for time:

48 = (35 * t) + (1/2) * (3 * t^2)

Simplifying the equation:

48 = 35t + (3/2) * t^2
0 = (3/2) * t^2 + 35t - 48

To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

In our equation, a = 3/2, b = 35, and c = -48. Plug these values into the quadratic formula:

t = (-35 ± sqrt(35^2 - 4 * (3/2) * (-48))) / (2 * (3/2))

Now, we can simplify and solve for t:

t = (-35 ± sqrt(1225 + 288)) / 3
t = (-35 ± sqrt(1513)) / 3

Now, we have two possible values of t. However, since the car is accelerating in the same direction as the train, we consider the positive root:

t = (-35 + sqrt(1513)) / 3

Using a calculator, we can evaluate this to find the value of t:

t ≈ 2.58 seconds

Now, we have the time it takes for the car to cover the initial distance. To find the speed of the car just as it passes the train, we can use the equation of motion:

final velocity = initial velocity + acceleration * time

Plugging in the values:

final velocity = 35 + (3 * 2.58)
final velocity ≈ 35 + 7.74
final velocity ≈ 42.74 m/s

Therefore, the speed of the car just as it passes the train is approximately 42.74 m/s.