During a race, one runner runs at a 6 min/mi pace and another runner runs at a 7 min/mi pace. The fast runner passes the slow runner. After a 2 second delay, the slow runner initiates an acceleration, catching up with the fast runner in 968 feet. Give this acceleration in ft/sec^2.
I have figured out that after the 2 seconds when the fast runner passes the slow runner, there are 4.19047 feet between the runners. I got this by:
5280 ft/360 sec = x ft/2 sec and
5280 ft/420 sec = x ft/2 sec then finding the difference in the distance the runners had travelled over those 2 seconds. However, I can't figure out what to do now.
To find the acceleration in ft/sec^2, you'll need to use the kinematic equations of motion. We can start by finding the initial velocities of both runners.
Let's assume the fast runner starts at position x = 0 ft and the slow runner starts at position x = 4.19047 ft (the distance between them after 2 seconds).
Using the formula for uniform motion:
distance = velocity * time
For the fast runner:
4.19047 ft = (6 min/mi) * (1 mi/5280 ft) * t sec
Simplifying, we get:
t = (4.19047 ft) / (6 min/mi * 1 mi/5280 ft) = 0.132651 sec
So after 0.132651 seconds, the fast runner has caught up to the slow runner.
Now, let's consider the time it takes for the slow runner to catch up to the fast runner:
distance = velocity * time
968 ft = (7 min/mi) * (1 mi/5280 ft) * t sec
Simplifying, we get:
t = (968 ft) / (7 min/mi * 1 mi/5280 ft) = 0.276591 sec
Now, we can calculate the acceleration using the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
For the slow runner:
968 ft = 0 ft/sec * 0.276591 sec + (1/2) * a * (0.276591 sec)^2
Simplifying, we get:
a = (2 * 968 ft) / (0.076522 sec^2) = 50,324.6496 ft/sec^2
Therefore, the acceleration of the slow runner is approximately 50,324.6496 ft/sec^2.