Sure, I'd be happy to help and hopefully make you laugh along the way! To solve this problem, we can use the equation:
d = (v^2 - u^2) / (2a)
where:
d = distance covered
v = final velocity
u = initial velocity
a = acceleration
Here, the initial velocity, u, is 25.0 m/s, the distance covered, d, is 95.0 m, and the acceleration, a, is -3.75 m/s^2.
Plugging in the given values, we have:
95 = (0 - 25^2) / (2 * (-3.75))
Now, let's solve for the final velocity, v:
95 = (-625) / (-7.5)
95 = 625 / 7.5
95 = 83.33
Uh-oh, it seems I made a clownish mistake here. Let me try again!
To find the maximum reaction time, we can use the equation:
d = ut + (1/2)at^2
where:
d = distance covered
u = initial velocity
t = time taken
a = acceleration
Rearranging the equation, we get:
t^2 - (2u/a)t + (2d/a) = 0
Now, using the quadratic formula:
t = [-(2u/a) Β± β((2u/a)^2 - 4(2d/a)] / 2
Substituting the given values:
t = [-(2 * 25.0 / -3.75) Β± β((2 * 25.0 / -3.75)^2 - 4(2 * 95.0 / -3.75)] / 2
Simplifying the equation:
t = (2 * 25.0 / 3.75) Β± β((2 * 25.0 / 3.75)^2 + (4 * 2 * 95.0 / 3.75)) / 2
t = (50 / 3.75) Β± β(800 / 3.75) / 2
t = 13.33 Β± β213.33 / 2
Now, calculating the values:
t β 13.33 Β± 14.6 / 2
t β 13.33 + 14.6 / 2 or t β 13.33 - 14.6 / 2
t β 27.93 / 2 or t β -1.27 / 2
So, the maximum reaction time is approximately 13.97 seconds, and since you mentioned it should be less than 0.5 seconds, there might be some confusion or an error somewhere in the calculations.
Remember, always double-check calculations and watch out for those pesky negative signs!