In the lab, a student is investigating the effects of texting and driving on reaction time. A student determines his/her reaction time by having a friend release a meter stick at
random intervals while the driver sends text messages. The student, on average,
catches the meter stick after it has traveled 0.64 m.
How far will the student
travel in a car moving at 25 m/s horizontally before reacting?

9.0 m
0.64 m
69 m

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:

link

object 2 because all the other ones are wrong