David is driving at a steady 36.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 1.20 m/s2 at the instant when David passes. How far (in m) does Tina drive before passing David?
d = 36 t
d = (1/2)(1.2) t^2
so
36 t = 0.6 t^2
t = 36/.6
then d = 36 t so d = (36)^2/.6
To find the distance Tina drives before passing David, we need to determine the time it takes for Tina to catch up with David.
First, we need to determine the time it takes for Tina to reach David's speed. We can use the formula:
v = u + at
where:
v = final velocity (David's speed)
u = initial velocity (Tina's initial speed)
a = acceleration
t = time
Rearranging the formula to solve for time (t), we have:
t = (v - u) / a
Substituting the values:
v = 36.0 m/s (David's speed)
u = 0 m/s (Tina's initial speed)
a = 1.20 m/s² (Tina's acceleration)
t = (36.0 - 0) / 1.20
t = 30.0 s
Now that we know it takes Tina 30.0 seconds to reach David's speed, we can calculate the distance she travels during that time.
Using the equation:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (0 m/s)
t = time (30.0 s)
a = acceleration (1.20 m/s²)
s = 0(30.0) + (1/2)(1.20)(30.0)²
s = 0 + (1/2)(1.20)(900.0)
s = 0 + (0.6)(900.0)
s = 540.0 m
Tina drives a distance of 540.0 meters before passing David.