David is driving a steady 24.0m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.70m/s^2 at the instant when David passes.

How far does Tina drive before passing David?

What is her speed as she passes him?

see above.

32m

To find the distance Tina drives before passing David, we can first find the time it takes for her to catch up to him. Then we can use this time to calculate the distance.

Let's assume that Tina catches up to David after time 't'.

Using the first equation of motion:
v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

For Tina:
Her initial velocity, u = 0 m/s (since she is at rest).
Her final velocity, v = 24.0 m/s (same as David's speed).
Her acceleration, a = 2.70 m/s^2.

Using the equation for Tina:
24.0 = 0 + 2.70t

Simplifying the equation:
2.70t = 24.0

To find the time, solving for 't':
t = 24.0 / 2.70
t ≈ 8.89 seconds

Now that we know it takes Tina approximately 8.89 seconds to catch up to David, we can calculate the distance she drives before passing him.

Using the second equation of motion:
s = ut + 0.5at^2

Where:
s = distance
u = initial velocity
t = time
a = acceleration

For Tina:
Her initial velocity, u = 0 m/s (since she is at rest).
Her time, t = 8.89 seconds (calculated previously).
Her acceleration, a = 2.70 m/s^2.

Using the equation for Tina:
s = 0 + 0.5 * 2.70 * (8.89)^2
s ≈ 107.18 meters

Therefore, Tina drives approximately 107.18 meters before passing David.

To find Tina's speed as she passes David, we can use the first equation of motion:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration

In this case, Tina's initial velocity is 0 m/s (since she is at rest) and her acceleration is 2.70 m/s^2 (given).

Using the equation for Tina:
v = 0 + 2.70 * 8.89
v ≈ 23.97 m/s

Therefore, Tina's speed as she passes David is approximately 23.97 m/s.