Reaction time. Your reaction time can be measured with the help of a fried using a ruler. Your friend holds the ruler vertically between your thumb and index finger. When he releases the ruler, you grab it as soon as you can. If the ruler is placed with the \(0 \mathrm{~cm}\) mark initially between your fingers, how can you use how far the ruler has fallen to find your reaction time? You can assume that you use negligible time to actually grab the ruler as soon as you start moving your finger. (a) Draw a free-body diagram for the ruler when it is falling. (b) Find the position of the ruler as a function of time. (c) Find your reaction time, if the ruler fell a vertical distance \(h\) before you grabbed it. (d) If you are driving in your car at \(80 \mathrm{~km} / \mathrm{h}\), how far do you travel during your reaction time?

Short Answer

Expert verified
Use \( t = \sqrt{\frac{2h}{g}} \) to find reaction time. Distance traveled is \( 22.22 \ m/s \times t \).

Step by step solution

01

Draw a Free-Body Diagram

Sketch the ruler falling under the influence of gravity. Label the downward gravitational force acting on the ruler.
02

Position of the Ruler as a Function of Time

Use the equation for the distance fallen under constant acceleration. Given the initial height is zero and initial velocity is zero, use the formula: \[ h(t) = \frac{1}{2} g t^2 \] where \( h \) is the vertical distance fallen, \( g \) is the acceleration due to gravity (\( 9.81 \ m/s^2 \)), and \( t \) is the time.
03

Solving for Reaction Time

Rearrange the formula \( h = \frac{1}{2} g t^2 \) to solve for \( t \): \[ t^2 = \frac{2h}{g} \] \[ t = \sqrt{\frac{2h}{g}} \]
04

Calculate Distance Traveled by Car During Reaction

Convert your speed from km/h to m/s: \[ 80 \ km/h = \frac{80 \times 1000}{3600} \ m/s = 22.22 \ m/s \] Use the formula distance = speed \( \times \) time: \[ d = 22.22 \ m/s \times t \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free-Body Diagram
To understand how the ruler falls, we need to draw a free-body diagram. This diagram shows all the forces acting on the ruler. Since the ruler is falling freely, the primary force acting on it is gravity. Gravity pulls the ruler downward. In our diagram, we label this force as \( F_g \), and it points straight down.
When you sketch this, show the ruler with an arrow pointing downward. Label the arrow with \( F_g \). Since no other significant forces (like air resistance) are acting on it, gravity is the only force we need to consider.
Drawing a clear and concise free-body diagram helps visualize the problem, making future steps easier to handle.
Constant Acceleration
In this situation, the ruler is under constant acceleration due to gravity. Constant acceleration means the velocity of the ruler increases at a steady rate as it falls.
Gravity on Earth causes objects to accelerate at approximately \( 9.81 \ m/s^2 \). This value is denoted by the symbol \( g \).
Since the problem assumes negligible air resistance, the only acceleration we consider is due to gravity. This uniform acceleration can be described by the equation:
\[ h(t) = \frac{1}{2} \times g \times t^2 \]
Here, \( h(t) \) represents the distance the ruler falls over time \( t \). This equation shows how the distance fallen increases with the square of the time under a constant acceleration, which simplifies understanding how the ruler's fall progresses.
Distance-Time Relationship
The relationship between the distance fallen by the ruler and the time taken is quadratic. This is evident from the equation:
\[ h(t) = \frac{1}{2} \times g \times t^2 \]
To find your reaction time, we rearrange this equation to solve for \( t \). If the ruler falls a distance \( h \), we use:
\[ t^2 = \frac{2h}{g} \]
Taking the square root of both sides gives us:
\[ t = \sqrt{ \frac{2h}{g}} \]
This equation shows your reaction time in terms of the distance the ruler falls.
Understanding this relationship helps you see that the longer the ruler falls, the longer your reaction time. Conversely, a shorter fall distance means a quicker reaction.
Units Conversion
In physics problems, units matter a lot. We need to convert units to ensure consistency. For example, in this problem, speeds are given in km/h, but we need m/s for calculations.
To convert from km/h to m/s, use the conversion factor \[ 1 \text{ km/h} = \frac{1000 \text{ meters}}{3600 \text{ seconds}} \]
For 80 km/h, the conversion is done as follows:
\[ 80 \text{ km/h} = \frac{80 \text{ km} \times 1000}{3600} \text{ seconds} \ = 22.22 \text{ m/s} \]
This step is crucial for accurate calculations, like finding the distance traveled by your car during your reaction time. Using the converted speed, we calculate the distance as:
\[ d = 22.22 \text{ m/s} \times t \]
Such conversions ensure your equations and answers are accurate and meaningful.

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Most popular questions from this chapter

In the army. You are told by a friend in the army that the force you feel when you fire a gun is the same as the force felt by a sandbag hit by the bullet because the two forces are action-reactions pairs. Is this true?

Experiments in Pisa. On a visit to Pisa, you decide to redo Galileo's original experiment based on your knowledge of physics. You bring to steel spheres of the same size to the top of the tower. One sphere is hollow and the other is solid. (a) Draw a free-body diagram for one of the spheres. (b) How would you describe air resistance for each of the spheres? (c) Find an expression for the acceleration of the sphere as a function of its mass. (d) Which of the two spheres have the largest acceleration?

Forces on a drop of water. A drop of water is hanging from a faucet. (a) Identify the forces acting on the drop and draw a free-body diagram of the drop. The drop finally falls down towards the sink. (b) Identify the forces acting on the drop and draw a free-body diagram of the drop.

Two masses and a spring. Two particles of \(m=0.1 \mathrm{~kg}\) are attached with a spring with spring constant \(k=100 \mathrm{~N} / \mathrm{m}\) and equilibrium length \(b=0.01 \mathrm{~m}\). Both particles start at rest and the spring is at equilibrium. An external force \(F=1000 \mathrm{~N}\) acts during \(1 \mathrm{~s}\) on one of the particles in the direction of the other particle. Find the position of both particles as a function of time from the time \(t=0 \mathrm{~s}\) when the external force starts acting. (You may solve this problem analytically or numerically).

Hammerhead. The head of your hammer is loose. How would you hit the shaft in order to fasten the hammerhead? Does this work if you are an astronaut working in space?

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