/*! 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 28 When striking, the pike, a preda... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When striking, the pike, a predatory fish, can accelerate from rest to a speed of \(4.0 \mathrm{m} / \mathrm{s}\) in \(0.11 \mathrm{s}\) a. What is the acceleration of the pike during this strike? b. How far does the pike move during this strike?

Short Answer

Expert verified
a. The acceleration of the pike during its strike is \(36.36 m/s^2\). b. The pike moves approximately \(0.2 m\) during this strike.

Step by step solution

01

Calculate the acceleration

Start by using the definition of acceleration (\(a\)) which is change in speed (\(v\)) over time (\(t\)). Here, it is given that the pike can accelerate from rest, means that it started with 0 m/s and achieved a speed of 4 m/s in 0.11 seonds. So, the formula becomes: \(a = \frac{v - u}{t}\) Here, \(v\) is final velocity, \(u\) is initial velocity and \(t\) is time taken. Substituting the given values ( \(v = 4 \ m/s, \ u = 0 \ m/s, \ t = 0.11 \ s\) ) into the formula, then evaluate the right-hand side.
02

Calculate the distance covered

The distance (\(s\)) covered by the pike can be calculated using the formula: \(s = ut + \frac{1}{2}at^2\) Here, \(u\) is initial velocity, \(a\) is acceleration just calculated, \(t\) is time and \(s\) is the covered distance. Again, considering that the initial velocity (\(u\)) is 0 m/s, the equation simplifies to: \(s = 0 \times t + \frac{1}{2}at^2 = \frac{1}{2}at^2\) Substitute in the values for acceleration (\(a\)) from Step 1 and time (\(t\)) from the question and then evaluate the expression to find the distance. This will provide the distance covered by the pike during the strike.

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.

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It provides us with the tools to describe an object's position, velocity, and acceleration over time. The most fundamental aspects of kinematics include understanding how to describe displacement, velocity (both average and instantaneous), and acceleration.

To paint a clearer picture, let's consider a pike, a swift aquatic predator. When analyzing its striking motion from rest to a certain speed, we are dealing with a straight-line motion known as rectilinear kinematics. In our example, the pike starts at rest and suddenly leaps forward to capture its prey, which is a classic case of rectilinear motion where we can apply kinematic principles to determine the pike’s velocity and the distance covered during its strike.
Motion Equations
Motion equations, often referred to as the equations of motion, are mathematical expressions that predict the future position of an object moving under the influence of uniform acceleration. They bind together an object's initial velocity, final velocity, acceleration, time, and displacement in coherent formulas.

For instance, in our exercise, the motion equation for acceleration calculates how quickly the pike can increase its speed during its strike. Another equation helps us find the distance covered by the pike while accelerating. These equations of motion are powerful as they provide a quantitative understanding of motion that can be applied to a wide range of scenarios, beyond our aquatic hunter here.
Uniform Acceleration
Uniform acceleration occurs when an object's speed changes at a constant rate over time. This is an essential concept in kinematics because it greatly simplifies the motion analysis. When acceleration is constant, we can use standard formulas to calculate various parameters of the motion.

In our example with the pike, we assume that the acceleration is uniform while it strikes. With uniform acceleration, the pike increases its speed from zero to 4.0 m/s consistently over 0.11 seconds. This constant rate of acceleration is what allows us to use straightforward motion equations to calculate the pike's acceleration and the distance it moves during the strike, illustrating how uniform acceleration makes solving kinematic problems more manageable.

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

Richard is driving home to visit his parents. 125 mi of the trip are on the interstate highway where the speed limit is 65 mph. Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at \(70 \mathrm{mph} .\) How many minutes does he save?

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched \(9.0 \mathrm{m}\) above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him. If he left the trampoline with a speed of \(8.0 \mathrm{m} / \mathrm{s}\), how long is he in the air before he catches the ball?

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at \(5.0 \mathrm{m} / \mathrm{s}\). Once free of this area, it speeds up to \(12 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{s}\). At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final speed?

A truck driver has a shipment of apples to deliver to a destination 440 miles away. The trip usually takes him 8 hours. Today he finds himself daydreaming and realizes 120 miles into his trip that he is running 15 minutes later than his usual pace at this point. At what speed must he drive for the remainder of the trip to complete the trip in the usual amount of time?

You are driving to the grocery store at \(20 \mathrm{m} / \mathrm{s}\). You are 110 \(\mathrm{m}\) from an intersection when the traffic light turns red. Assume that your reaction time is \(0.70 \mathrm{s}\) and that your car brakes with constant acceleration. a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop?

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.