We know that we can write the speed of something as the quotient between the distance it moves and the time it takes to move that distance.
speed = distance/time
Using this, we will find that the correct option is the first one, 9.0 m
The information we know is:
The student, on average, catches the meter stick after it has traveled 0.64m
We need to transform this into time.
Assuming that the meter stick is just released, the only acceleration it will have is the gravitational one, so we have:
a(t) = -9.8m/s^2
To get the velocity of the stick we need to integrate over time, as the it is just released, it will not have initial velocity, thus the constant of integration is zero:
v(t) = (-9.8m/s^2)*t
To get the position equation we need to integrate again, and as the initial position is relative, we can define it as zero, so again the constant of integration is zero:
p(t) = (1/2)(-9.8m/s^2)*t^2
p(t) = (-4.9 m/s^2)*t
We know that the student catches the stick after it traveled (downwards) 0.64m
Then we need to solve:
-0.64m = (-4.9 m/s^2)*t
√[(-0.64m)/(-4.9m/s^2)] = t = 0.36s
So the reaction time of the student is 0.36 seconds.
Now we can use the first equation:
speed = distance/time
And rewrite it as:
distance = speed*time
So if the car moves at 25m/s and the student needs 0.36 s to react, the distance that he/she will move in that time frame is:
d = (25m/s)*0.36s = 9.0m
So the correct option is the first one.
If you want to learn more, you can read:
https://brainly.com/question/12550364
