/*! 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 38 Chameleons catch insects with th... [FREE SOLUTION] | 91Ó°ÊÓ

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Chameleons catch insects with their tongues, which they can rapidly exlend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable \(250 \mathrm{m} / \mathrm{s}^{2}\) for \(20 \mathrm{ms}\), then travels at constant speed for another \(30 \mathrm{ms}\). During this total time of \(50 \mathrm{ms}, 1 / 20\) of a second, how far does the tongue reach?

Short Answer

Expert verified
The tongue of the chameleon reaches a total of \(0.2 \mathrm{m}\).

Step by step solution

01

Calculate Initial Velocity

First, calculate the velocity attained by the tongue during the acceleration phase. This is done using the formula \(v = at\), where \(a = 250 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration and \(t = 20 \mathrm{ms} = 0.02 \mathrm{s}\) is the time. This gives us \(v = 250*0.02 = 5 \mathrm{m/s}\).
02

Calculate Acceleration Phase Distance

Next, calculate the distance travelled during the acceleration phase using the formula \(d = \frac{1}{2}at^2\). This gives us \(d=0.5*250*0.02^2 = 0.05 \mathrm{m}\).
03

Calculate Constant Speed Phase Distance

Calculate the distance travelled during the constant speed phase using the formula \(d = vt\), where \(v = 5 \mathrm{m/s}\) from Step 1 and \(t = 30 \mathrm{ms} = 0.03 \mathrm{s}\). This gives us \(d = 5*0.03 = 0.15 \mathrm{m}\).
04

Calculate Total Distance

Add the distances travelled during the acceleration and constant speed phases to obtain the total distance travelled by the tongue. This gives us \(0.05 \mathrm{m} + 0.15 \mathrm{m} = 0.2 \mathrm{m}\).

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Key Concepts

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

Constant Acceleration
Understanding constant acceleration is crucial when studying the motion of objects in physics. Constant acceleration occurs when an object speeds up or slows down at a steady rate over time. The example of the chameleon's tongue in the exercise exhibits this behavior initially, as it accelerates at a rate of 250 m/s^2 for 20 milliseconds.

Imagine you're driving a car that goes faster by the same amount each second; that's constant acceleration. In physics problems, when an object is known to have constant acceleration, specific formulas called kinematic equations come into play. These equations relate the object's initial velocity, final velocity, acceleration, time, and displacement.

Importance of Constant Acceleration

Constant acceleration allows us to predict an object's position and velocity at any given time. In the chameleon's case, knowing the acceleration and time helps us calculate how fast the tongue reaches its peak speed. It's a concept that not only has applications in biology but also in engineering, aeronautics, and many other fields.
Motion Formulas
Motion formulas, also known as kinematic equations, are a set of four equations that predict the future movement of an object based on its current state of motion. These equations assume constant acceleration and are derived from the basic principles of motion.

For our chameleon, two motion formulas were particularly useful:
  • For velocity: \(v = at\), to figure out the chameleon tongue's velocity after the acceleration phase.
  • For distance during acceleration: \(d = \frac{1}{2}at^2\), to calculate how far the tongue extends while it is speeding up.
With these formulas, we could determine the tongue's speed at the end of the constant acceleration period and the distance it traveled in that time. It's like solving a mystery where you piece together clues (initial velocity, acceleration, and time) to find out where the object ends up and how quickly it gets there.
Velocity-Time Calculations
Velocity-time calculations are at the heart of kinematic motion problems. They help us understand how an object's velocity changes with time and can be used to calculate distance traveled. After finding the velocity at the end of the acceleration phase, as we did with the chameleon's tongue, we can determine the distance traveled during the constant velocity phase.

The formula \(d = vt\) is pivotal here since the tongue moves at a constant velocity (5 m/s) for a particular time (0.03 s). It is the simplicity of this part of the motion that contrasts with the need to account for acceleration in earlier phases. Through these calculations, students can appreciate the varying complexities of motion and hone their problem-solving skills by applying the right formula at the right time.

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Most popular questions from this chapter

You're driving down the highway late one night at \(20 \mathrm{m} / \mathrm{s}\) when a deer steps onto the road \(35 \mathrm{m}\) in front of you. Your reaction time before stepping on the brakes is \(0.50 \mathrm{s},\) and the maximum deceleration of your car is \(10 \mathrm{m} / \mathrm{s}^{2}\). a. How much distance is between you and the deer when you come to a stop? b. What is the maximum speed you could have and still not hit the deer?

A simple model for a person running the \(100 \mathrm{m}\) dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of \(11.2 \mathrm{m} / \mathrm{s}\) in \(2.14 \mathrm{s}\), what will be his total time?

A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is \(1 / 4\) mile long. A large, fully loaded passenger jet has about the same acceleration from rest, but it needs to achieve twice the speed to take off. What is the minimum runway length that will serve? Hint: You can solve this problem using ratios without having any additional information.

A football is kicked straight up into the air; it hits the ground 5.2 s later. a. What was the greatest height reached by the ball? Assume it is kicked from ground level. b. With what speed did it leave the kicker"s foot?

II Haley is driving down a straight highway at 75 mph. A construction sign warns that the speed limit will drop to \(55 \mathrm{mph}\) in \(0.50 \mathrm{mi} .\) What constant acceleration (in \(\mathrm{m} / \mathrm{s}\) ) will bring Haley to this lower speed in the distance available?

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