John is driving at a constant speed of 20.0 m/s when he passes Jane, who is traveling in the same direction at 15.0 m/s at the moment that John passes her. If Jane is accelerating at 2.0 m/s2, how far will Jane drive before she catches up to John? How much time passes from when John passes Jane until Jane catches back up with John? (You can assume all motion is on flat, level ground.)

John distance=20*time

Jane distance=15*time+1/2 a time^2

set them equal, solve for time
then solve for either distance in that time.

Why the 1/2 in 1/2*a*time^2?

To find the distance Jane will drive before she catches up to John, we can set up an equation based on their relative speeds and the time it takes for Jane to catch up.

Let's call the distance Jane drives before she catches up as "D" and the time it takes for Jane to catch up as "t".

Now, let's break down the problem into steps:

Step 1: Find the acceleration time for Jane.
We can use the formula: v = u + at, where
v = final velocity (Jane's velocity when she catches up to John)
u = initial velocity (Jane's velocity when John passes her)
a = acceleration (constant for Jane)

Plugging in the given values:
v = 20.0 m/s (John's speed),
u = 15.0 m/s (Jane's speed when John passes her),
a = 2.0 m/s^2 (Jane's acceleration)

v = u + at
=> 20.0 = 15.0 + 2.0t

Simplifying the equation:
2.0t = 5.0
t = 5.0 / 2.0
t = 2.5 seconds

Step 2: Use the time (t) to find Jane's distance (D) using the equation: d = ut + 1/2at^2, where
d = distance (Jane's distance when she catches up to John)
u = initial velocity (Jane's velocity when John passes her)
t = time
a = acceleration (constant for Jane)

Plugging in the given values:
u = 15.0 m/s (Jane's speed when John passes her),
t = 2.5 seconds (time it takes for Jane to catch up),
a = 2.0 m/s^2 (Jane's acceleration)

d = ut + 1/2at^2
=> d = (15.0)(2.5) + 1/2(2.0)(2.5)^2

Simplifying the equation:
d = 37.5 + 1/2(2.0)(6.25)
d = 37.5 + 1/2(12.5)
d = 37.5 + 6.25
d = 43.75 meters

Therefore, Jane will drive approximately 43.75 meters before catching up to John.

To find how much time passes from when John passes Jane until Jane catches back up with John, we already calculated that it takes Jane 2.5 seconds to catch up to John.