David is driving a steady 21.0m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.50m/s2 at the instant when David passes.
How far does Tina drive before passing David?
What is her speed as she passes him?
david distance: 21*time
tina distance: 1/2 a time^2
they both go the same distance, in the same time. set them equal, solve for time. Then calculate distance.
Her final speed=a*time
115.2
To answer the question, we need to analyze the motion of both David and Tina.
First, we can calculate the time it takes for Tina to catch up to David. We can use the equation of motion:
π = π£0π‘ + (1/2)ππ‘^2
Where,
π = displacement
π£0 = initial velocity
π = acceleration
π‘ = time
For David, his initial velocity is 21.0 m/s, and since he maintains a steady speed, his acceleration is 0. Hence, the equation simplifies to:
π = 21.0π‘
For Tina, her initial velocity is 0, and her acceleration is 2.50 m/s^2. The equation becomes:
π = (1/2) * 2.50 * π‘^2
To find the time it takes for Tina to catch up to David, we can equate the two equations and solve for π‘:
21.0π‘ = (1/2) * 2.50 * π‘^2
Let's solve this equation to find the value of π‘.