In order to pass a physical education class at a university, a student must run 1.0 mi in 12 min. After running for 10 min, she still has 500 yd to go. If her maximum acceleration is 0.15 m/s2, can she make it? If the answer is no, determine what acceleration she would need to be successful.
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yes
To determine whether the student can make it or not, we need to compare the distance she can cover in the remaining time to the distance she still needs to run.
Let's start by converting all the units to a consistent system. Since her running time and distance are given in minutes and yards, we need to convert them to seconds and meters, respectively, to match the acceleration unit.
1 mile is equal to 1760 yards, so 1.0 mile is equal to 1.0 * 1760 = 1760 yards.
The remaining distance in yards is 500.
1 yard is equal to 0.9144 meters, so the remaining distance in meters is 500 * 0.9144 = 457.2 meters.
The given maximum acceleration is 0.15 m/s^2.
Now, let's determine the distance the student can cover in the remaining time.
Given that she has already run for 10 minutes, there are 12 - 10 = 2 minutes remaining.
Converting the remaining time to seconds gives us 2 * 60 = 120 seconds.
We can use the kinematic equation to find the distance covered using the given maximum acceleration:
d = v0*t + 0.5*a*t^2,
Where:
d is the distance covered,
v0 is the initial velocity (0 m/s since the student starts from rest),
a is the acceleration,
t is the time.
Substituting the values we have:
d = 0 + 0.5 * 0.15 * (120)^2 = 1080 meters.
Now, let's compare the distance she can cover (1080 meters) to the distance she still needs to run (457.2 meters).
Since 1080 meters is greater than 457.2 meters, she can make it to the finish line within the remaining time.
Therefore, the answer is YES, she can make it.
If the answer was NO, we would need to determine the acceleration she would need to be successful. We would use the same kinematic equation to find the required acceleration (a) to cover the remaining distance (457.2 meters) in the remaining time (120 seconds). Rearranging the equation:
a = 2 * (d - v0*t) / t^2,
Where:
a is the acceleration,
d is the remaining distance,
v0 is the initial velocity (0 m/s since the student starts from rest),
t is the time.
Substituting the values we have:
a = 2 * (457.2 - 0) / (120)^2 = 0.015 m/s^2.
Therefore, if the student couldn't make it, she would need an acceleration of 0.015 m/s^2 to successfully complete the remaining distance in the given time.
so she has 500 yards to go, 2 min, her velocity at start is 1200yards/10 min
Vf^2=Vi^2+2ad
So vf=vi+at
(vi+at)^2=vi^2 + ad
Determine vi more accurately than I did above, convert it to m/s, then solve for t given max a.
If no, then put t in as 120sec, and solve for a.