KEY2PHYSICS

Problem : / 1

#IB-272

Reaction Time

MAKING CONNECTIONS: TAKE-HOME EXPERIMENT - REACTION TIME

A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?

EXAMPLE 1 CALCULATING VELOCITY OF A FALLING OBJECT: A ROCK THROWN DOWN

What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.

Strategy

Draw a sketch.

Figure 1

Since up is positive, the final position of the rock will be negative because it finishes below the starting point at \(y_{0}=0.\) Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.

Solution

1. Identify the knowns. \(y_0=0;\ y_1=−5.10 m; \ v_0=−13.0 m/s; \ a=−g=−9.80 m/s^2.\)

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation \(v^2=v_0^2-2g(y−y0)\) works well because the only unknown in it is \(v\). (We will plug \(y_1\) in for \(y\).)

3. Enter the known values

\(v_2=(−13.0\; m/s)^2+2(−9.80 \;m/s^2)(−5.10 \;m−0 m)=268.96 \;m^2/s^2,\)

2.80

where we have retained extra significant figures because this is an intermediate result.
Taking the square root, and noting that a square root can be positive or negative, gives

\(v=\pm16.4\; m/s\)

2.81

The negative root is chosen to indicate that the rock is still heading down. Thus,

\(v=-16.4 \;m/s\)

2.82

Discussion

Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 1 and Figure 2(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of \(\pm3.20\; m/s\) is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.

Figure 2 (a) A person throws a rock straight up, as explored in Example 1. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2. Note that at the same distance below the point of release, the rock has the same velocity in both cases.

Another way to look at it is this: In Example 1, the rock is thrown up with an initial velocity of 13.0 \(\mathrm{m/s}\). It rises and then falls back down. When its position is \(y=0\) on its way back down, its velocity is −13.0 \(\mathrm{m/s}\). That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of \(y=−5.10 \thinspace \mathrm{m}\) to be the same whether we have thrown it upwards at +13.0 \(\mathrm{m/s}\) or thrown it downwards at −13.0 \(\mathrm{m/s}\). The velocity of the rock on its way down from \(y=0\) is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.

EXAMPLE 2 FIND G FROM DATA ON A FALLING OBJECT

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 3. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Figure 3 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?

Strategy

Draw a sketch.

Figure 4

We need to solve for acceleration \(a\). Note that in this case, displacement is downward and therefore negative, as is acceleration.

Solution

1. Identify the knowns. \(y_{0}=0; y=–1.0000 \thinspace \mathrm{m}\); \(t=0.45173\); \(v_{0}=0\).

2. Choose the equation that allows you to solve for \(a\) using the known values.

\(y = y_{0}+v_{0}t+\frac{1}{2}at^2\)

3. Substitute 0 for \(v_{0}\) and rearrange the equation to solve for \(a\). Substituting 0 for \(v_{0}\) yields

\(y = y_{0}+ \frac{1}{2}at^2\)

Solving for \(a\) gives

\(a=\frac{2(y-y_{0})}{t^2}\)

4. Substitute known values yields

\(a = \frac{2(-1.0000 \thinspace \mathrm{m-0})}{ (0.45173 \thinspace \mathrm{s}) ^2 } = -9.8010 \thinspace \mathrm{m/s^2}\)

so, because \(a=-g\) with the directions we have chosen,

\(g=9.8010 \thinspace \mathrm{m/s^2}\)

Discussion

The negative value for \(a\) indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of , so makes sense. Since the data going into the calculation are relatively precise, this value for \(g\) is more precise than the average value of ; it represents the local value for the acceleration due to gravity.(-)