/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A meter stick is held vertically... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) \(t = \sqrt{\frac{2d}{g}}\); (b) the reaction time is approximately 0.19 seconds.

Step by step solution

01

Understand the Motion

The meter stick is in free fall until you catch it, so it undergoes uniform acceleration due to gravity. The relevant equation for the motion is the kinematic equation: \[ d = \frac{1}{2} g t^2 \] where \(d\) is the distance fallen, \(g\) is the acceleration due to gravity (approximately \(9.8\, \text{m/s}^2\) on the surface of Earth), and \(t\) is the time it takes for the stick to fall (your reaction time).
02

Solve for Reaction Time

We need to solve the kinematic equation for the reaction time \(t\). Starting with \[ d = \frac{1}{2} g t^2 \]we can rearrange to find \(t\):\[ t^2 = \frac{2d}{g} \]Taking the square root of both sides gives us:\[ t = \sqrt{\frac{2d}{g}} \]
03

Substitute Known Values

Now, substitute the known values into the equation. The distance \(d\) is given as \(17.6\, \text{cm}\), which is \(0.176\, \text{m}\), and \(g = 9.8\, \text{m/s}^2\). Substitute these values into the equation for \(t\):\[ t = \sqrt{\frac{2 \times 0.176}{9.8}} \]
04

Calculate

Calculate the value of the reaction time:\[ t = \sqrt{\frac{0.352}{9.8}} \approx \sqrt{0.0359} \approx 0.1895 \, \text{seconds} \]
05

Conclusion

The reaction time when the meter stick falls 17.6 cm before being caught is approximately \(0.19\, \text{seconds}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental tools in physics that describe the motion of objects. They connect an object's displacement, velocity, acceleration, and time. These equations are especially useful in scenarios where an object moves with constant acceleration. In our example of a meter stick in free fall, we use these equations to calculate reaction time.
A core kinematic equation used here is:
  • \[ d = \frac{1}{2} g t^2 \]
This relates the distance (\(d\) ) the meter stick falls to the time (\(t\)) it takes, incorporating the acceleration due to gravity (\(g\)). By knowing the distance an object falls, we can solve for the time, which corresponds to your reaction time in this scenario. This is possible because the distances fallen in free fall scenarios can be calculated by understanding how gravity acts on the object over time.
To find the reaction time, rearrange the kinematic equation to solve for \(t\):
  • \[ t = \sqrt{\frac{2d}{g}} \]
Here, you simply plug in your known values of \(d\) and \(g\) to get \(t\).
Free Fall Motion
Free fall motion describes any motion of an object where gravity is the only force acting upon it. This means there's no air resistance or other force involved. In the exercise, the meter stick experiences free fall from the moment it is released until you catch it. During free fall, all objects accelerate at the same rate, regardless of their mass. Provided they are near the Earth's surface, they will accelerate at approximately \(9.8 \, \text{m/s}^2 \).
To fully grasp the motion of a falling object, such as our meter stick, it's important to consider the initial conditions. When you release the meter stick, its initial velocity is zero. The only influence on its motion is gravity, which increases its velocity as it falls. The further it falls before being caught, the more time has elapsed, allowing for a direct measure of reaction time.
This behavior is elegantly modeled by kinematic equations, allowing us to use measured distances to calculate reaction times. With a free-falling object, knowing just the distance fallen and the acceleration due to gravity is sufficient to compute the time of fall.
Acceleration due to Gravity
Acceleration due to gravity is the rate at which any object accelerates when it's in free fall at the Earth's surface. It's typically approximated as \(9.8 \, \text{m/s}^2 \). This acceleration is a constant, which means every second, the velocity of a freely falling object increases by \(9.8 \, \text{m/s}\), provided the object is falling directly downwards without any interference from other forces.
Gravity is a force that attracts two bodies towards each other, and on Earth, it gives mass weight. It acts as a central player in our exercise, the primary force that causes the meter stick to fall when released. Without this predictable acceleration, calculating things like reaction time from fallen objects would not be feasible.
When performing calculations involving gravity, it is crucial to ensure that other forces, such as air resistance, are negligible. This lets us assume that gravitational acceleration remains constant, simplifying our calculations significantly. In practical terms, this constant acceleration allows us to predict and understand the movements of falling objects, and when combined with kinematic equations, it provides a reliable way to solve for unknown variables like time or distance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. Ignore air resistance. (a) If the height of the building is 20.0 m, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_0\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_0\) is 6.0 m/s and (ii) \(v_0\) is 9.5 m/s? (c) If \(v_0\) is greater than some value \(v_{max}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{max}\). The value \(v_{max}\) has a simple physical interpretation. What is it? (d) If \(v_0\) is less than some value \(v_{min}\), no value of h exists that allows both balls to hit the ground at the same time. Solve for \(v_{min}\). The value \(v_{min}\) also has a simple physical interpretation. What is it?

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin at the nest and extend the +\(x\)-axis to the release point, what was the bird’s average velocity in m/s (a) for the return flight and (b) for the whole episode, from leaving the nest to returning?

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 960 m above the earth's surface. The rocket's engines give the rocket an upward acceleration of 16.0 m/s\(^2\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.