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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.

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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.

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Most popular questions from this chapter

A car is traveling at a steady \(80 \mathrm{km} / \mathrm{h}\) in a \(50 \mathrm{km} / \mathrm{h}\) zone. A police motorcycle takes off at the instant the car passes it, accelerating at a steady \(8.0 \mathrm{m} / \mathrm{s}^{2}\) a. How much time elapses before the motorcycle is moving as fast as the car? b. How far is the motorcycle from the car when it reaches this speed?

I Starting at \(48th\) Street, Dylan rides his bike due east on Meridian Road with the wind at his back. He rides for 20 min at \(15 \mathrm{mph} .\) He then stops for \(5 \mathrm{min},\) turns around, and rides back to 48th Street; because of the headwind, his speed is only 10 mph. a. How long does his trip take? b. Assuming that the origin of his trip is at \(48th\) Street, draw a position- versus-time graph for his trip.

A Thomson's gazelle can reach a speed of \(13 \mathrm{m} / \mathrm{s}\) in \(3.0 \mathrm{s}\). A lion can reach a speed of \(9.5 \mathrm{m} / \mathrm{s}\) in \(1.0 \mathrm{s}\). A trout can reach a speed of \(2.8 \mathrm{m} / \mathrm{s}\) in \(0.12 \mathrm{s}\). Which animal has the largest acceleration?

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?

Does a real automobile have con-stant acceleration? Measured data for a Porsche 944 Turbo at maximum acceleration are as shown in the table. a. Convert the velocities to \(\mathrm{m} / \mathrm{s},\) then make a graph of velocity versus time. Based on your graph, is the acceleration constant? Explain. b. Estimate how far the car traveled in the first \(10 \mathrm{s}\). c. Draw a smooth curve through the points on your graph, then use your graph to estimate the car"s acceleration at \(2.0 \mathrm{s}\) and \(8.0 \mathrm{s} .\) Give your answer in \(\mathrm{SI}\) units. Hint: Remember that acceleration is the slope of the velocity graph.$$ \begin{array}{cc} \mathbf{t}(\mathrm{s}) & \boldsymbol{v}_{\boldsymbol{x}}(\mathrm{mph}) \\ \hline 0 & 0 \\ 2 & 41 \\ 4 & 66 \\ 6 & 83 \\ 8 & 97 \\ 10 & 110 \\ \hline \end{array} $$

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