David is driving at a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.70 m/s^2 at the instant David passes. How far does Tina drive before passing David and what is her speed as she passes him?
Let t=0 when David passes Tina.
Consider the speed of Tina.
v0=0 m/s
a=2.7 m/s²
Assuming Tina accelerates indefinitely, at time t when she overtakes David, they would have travelled the same distance since t=0.
Distance travelled by David
= 30t
Distance travelled by Tina
=v0*t + (1/2)at²
Equate and solve for t.
Her speed would be a*t.
690 m
To find how far Tina drives before passing David, we need to determine the time it takes for Tina to catch up with him. We can use the equation:
\[ \text{Final Velocity} = \text{Initial Velocity} + \text{Acceleration} \times \text{Time} \]
For David's motion, his initial velocity is 30.0 m/s, and since there is no acceleration, his final velocity remains constant at 30.0 m/s.
For Tina's motion, her initial velocity is 0 m/s, and her acceleration is 2.70 m/s^2. We need to find the time it takes for Tina to reach a velocity of 30.0 m/s.
\[ 30.0 \, \text{m/s} = 0 \, \text{m/s} + 2.70 \, \text{m/s}^2 \times \text{Time} \]
\[ \text{Time} = \frac{30.0 \, \text{m/s}}{2.70 \, \text{m/s}^2} \]
Solving this equation, we find that Time is approximately 11.11 seconds.
Now that we know the time Tina takes to catch up with David, we can calculate the distance she travels during this time. We can use the equation:
\[ \text{Distance} = \text{Initial Velocity} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 \]
For Tina's motion:
\[ \text{Distance} = 0 \, \text{m/s} \times 11.11 \, \text{s} + \frac{1}{2} \times 2.70 \, \text{m/s}^2 \times (11.11 \, \text{s})^2 \]
Simplifying this equation, we find that Tina travels approximately 166.90 meters before passing David.
Finally, to find Tina's speed as she passes David, we can use the equation:
\[ \text{Final Velocity} = \text{Initial Velocity} + \text{Acceleration} \times \text{Time} \]
Substituting the known values:
\[ \text{Final Velocity} = 0 \, \text{m/s} + 2.70 \, \text{m/s}^2 \times 11.11 \, \text{s} \]
Solving this equation, we find that Tina's speed as she passes David is approximately 30.0 m/s.
To find the distance Tina drives before passing David, we need to find the time it takes for Tina to catch up with David. First, we need to determine the time it takes for David to catch up to Tina.
Let's assume the time it takes for David to catch up to Tina is t seconds. During this time, David's velocity remains constant at 30.0 m/s, and he covers a distance given by:
distance covered by David = velocity of David × time = 30.0 m/s × t
Now, let's consider Tina's motion. Tina starts at rest and accelerates uniformly at 2.70 m/s^2 until she catches up with David. The distance covered by Tina can be calculated using the equations of accelerated motion.
We can use the following equation to calculate the distance covered by Tina:
distance covered by Tina = initial velocity × time + (1/2) × acceleration × time^2
Tina's initial velocity is 0 m/s, and the acceleration is 2.70 m/s^2. We don't know the time, so let's call it t'.
To find the time it takes for Tina to catch up with David, we need to equate their distances and solve for t:
30.0 × t = 0 × t' + (1/2) × 2.70 × t'^2
Simplifying this equation, we get:
30.0 t = 1.35 t'^2
Now, let's find t' in terms of t:
t' = √(30.0 t / 1.35)
Now that we have the value of t', we can calculate the distance Tina covers using the equation for distance covered by Tina:
distance covered by Tina = initial velocity × time + (1/2) × acceleration × time^2
distance covered by Tina = 0 × t' + (1/2) × 2.70 × t'^2
Putting the values, we get:
distance covered by Tina = (1/2) × 2.70 × ( √(30.0 t / 1.35) )^2
Simplifying further, we get:
distance covered by Tina = 1.35 × t
Finally, we can calculate Tina's speed as she passes David. Since she has caught up with him, her speed will be the same as David's, which is 30.0 m/s.
Therefore, Tina drives a distance of 1.35 times the time David takes to catch up to her before passing him. Her speed as she passes him is 30.0 m/s.