A car starts from rest with an acceleration of 1.87 m/s^2 at the same time that a second car moving with a velocity of 29.2 m/s passes it. How far does the first car move before it overtakes the second car?

To find out how far the first car moves before it overtakes the second car, we need to determine the time it takes for the first car to catch up to the second car.

Let's start by finding the time it takes for the first car to catch up with the second car. We'll use the equation of motion that relates acceleration, initial velocity, time, and displacement:

v = u + at

where:
v = final velocity (29.2 m/s for the second car)
u = initial velocity (0 m/s as the first car starts from rest)
a = acceleration (1.87 m/s^2 for the first car)
t = time

Rearranging the equation:

t = (v - u) / a

t = (29.2 m/s - 0 m/s) / 1.87 m/s^2
t ≈ 15.57 s

So, it takes approximately 15.57 seconds for the first car to catch up to the second car.

Next, we can calculate the distance the first car moves before it overtakes the second car. We'll use the equation of motion that relates displacement, initial velocity, time, and acceleration:

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

where:
s = displacement
u = initial velocity (0 m/s as the first car starts from rest)
t = time (15.57 s as calculated earlier)
a = acceleration (1.87 m/s^2)

Plugging in the values:

s = (0 m/s)(15.57 s) + (1/2)(1.87 m/s^2)(15.57 s)^2

s ≈ 144.3 m

Therefore, the first car moves approximately 144.3 meters before it overtakes the second car.