Question

2.39 a simple reaction - time test. a meter stick is held vertically above your hand, with the lower end between your thumb and first finger. when you see the meter stick released, you grab it with those two fingers. you can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. (a) derive a relationship for your reaction time in terms of this measured distance, d. (b) if the measured distance is 17.6 cm, what is your reaction time?

Explanation:

Step1: Identify the kinematic equation

The meter - stick is in free - fall, and the kinematic equation for motion under constant acceleration $y = y_0+v_0t+\frac{1}{2}at^2$. Here, the initial position $y_0 = 0$, the initial velocity $v_0 = 0$, and the acceleration $a = g=9.8\ m/s^2$. The distance the stick falls is $d$, so $y=-d$ (taking downwards as negative).
$-d = 0+0\times t-\frac{1}{2}gt^2$

Step2: Solve for reaction time $t$

From $-d =-\frac{1}{2}gt^2$, we can solve for $t$. First, multiply both sides by $- 1$ to get $d=\frac{1}{2}gt^2$. Then, rearrange the equation for $t$:
$t=\sqrt{\frac{2d}{g}}$

Step3: Substitute the given value of $d$

Given $d = 17.6\ cm=0.176\ m$ and $g = 9.8\ m/s^2$.
$t=\sqrt{\frac{2\times0.176}{9.8}}$
$t=\sqrt{\frac{0.352}{9.8}}\approx\sqrt{0.03592}\approx0.19\ s$

Answer:

(a) $t=\sqrt{\frac{2d}{g}}$; (b) $t\approx0.19\ s$