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(I) A sprinter accelerates from rest to \(9.00 \mathrm{~m} / \mathrm{s}\) in \(1.28 \mathrm{~s}\). What is her acceleration in (a) \(\mathrm{m} / \mathrm{s}^{2} ;\) (b) \(\mathrm{km} / \mathrm{h}^{2}\) ?

Short Answer

Expert verified
(a) 7.03 m/s²; (b) 91.1 km/h².

Step by step solution

01

Identify Given Values

The initial velocity (\(u\)) is given as 0 m/s because the sprinter starts from rest. The final velocity (\(v\)) is 9.00 m/s. The time (\(t\)) taken to reach this velocity is 1.28 s.
02

Use Acceleration Formula

We can calculate the acceleration using the formula: \[ a = \frac{v - u}{t} \]Substitute the given values: \(v = 9.00\) m/s, \(u = 0\) m/s, and \(t = 1.28\) s.\[ a = \frac{9.00 - 0}{1.28} \]
03

Calculate Acceleration in m/s²

Perform the calculation:\[ a = \frac{9.00}{1.28} \approx 7.03 \text{ m/s}^2 \]
04

Convert m/s² to km/h²

To convert acceleration from m/s² to km/h², use the conversion factor where 1 m/s² equals 12960 km/h² (since 1 m/s = 3.6 km/h), we multiply by \(3.6^2 = 12.96\):\[ a = 7.03 \times 12.96 \approx 91.1 \text{ km/h}^2 \]

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Key Concepts

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

Acceleration
Acceleration is a fundamental concept in kinematics, referring to the rate of change of velocity over a specified time period. Understanding acceleration involves knowing how fast an object's velocity changes. For the sprinter mentioned in the exercise, acceleration is calculated by assessing how quickly she speeds up from her starting velocity of rest (0 m/s) to her final velocity of 9.00 m/s over 1.28 seconds. The formula to determine acceleration is:- \( a = \frac{v - u}{t} \) - \( v \) = final velocity - \( u \) = initial velocity, which is 0 m/s if starting from rest - \( t \) = time duration during which the velocity change occursIn this exercise, substituting the known values into the equation gives us:\[ a = \frac{9.00 - 0}{1.28} = 7.03 \, \text{m/s}^2 \]Understanding this concept is essential, as it explains how rapidly a sprinter can reach a certain speed.
Velocity
Velocity is a vector quantity that provides information about the speed and direction of an object's motion. In kinematics, it is crucial to distinguish between velocity and speed because velocity accounts for direction, while speed does not.For the sprinter, her final velocity is given as 9.00 m/s, which means she moves at this rate in a specific direction after accelerating from rest. Velocity is expressed in meters per second (m/s) in the SI unit system.To fully understand velocity, consider these points:- It indicates the rate of change of position.- Velocity is significant when analyzing an object's motion since it incorporates direction.- The initial velocity \( u \) here is 0 m/s since the sprinter starts from rest.With these basics, you can better understand how athletes achieve high velocities in short times through rapid acceleration.
Unit Conversion
Unit conversion is a mathematical method used to change the units of a measurement without altering its value. In physics, especially kinematics, it is often necessary to convert between different units to maintain consistency or to interpret results in more familiar units. In the given exercise, the acceleration calculated in meters per second squared (m/s²) needs to be converted to kilometers per hour squared (km/h²). The conversion process involves understanding the relationship between these units:- 1 m/s = 3.6 km/hFor acceleration, converting from m/s² to km/h² involves squaring the conversion factor to correctly apply it to the squared units:- \(1 \text{ m/s}^2 = 12.96 \text{ km/h}^2\)To perform this conversion:- Multiply the acceleration in m/s² by 12.96.Applying it to our calculated acceleration, we get:\[ 7.03 \times 12.96 = 91.1 \, \text{km/h}^2 \]Always ensure your conversions are precise to avoid misconstruing the data during analysis or discussion.

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Most popular questions from this chapter

(II) Determine the stopping distances for an automobile with an initial speed of \(95 \mathrm{~km} / \mathrm{h}\) and human reaction time of \(1.0 \mathrm{~s}:(a)\) for an acceleration \(a=-5.0 \mathrm{~m} / \mathrm{s}^{2} ; \quad\) (b) for \(a=-7.0 \mathrm{~m} / \mathrm{s}^{2}\)

(II) A car traveling at \(105 \mathrm{~km} / \mathrm{h}\) strikes a tree. The front end of the car compresses and the driver comes to rest after traveling \(0.80 \mathrm{~m} .\) What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of " \(g\) 's," where \(1.00 g=9.80 \mathrm{~m} / \mathrm{s}^{2}\).

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a \(9.0-\mathrm{km}\) trip in two situations: \((a)\) the stations at which the trains must stop are \(1.8 \mathrm{~km}\) apart (a total of 6 stations, including those at the ends); and (b) the stations are \(3.0 \mathrm{~km}\) apart (4 stations total). Assume that at each station the train accelerates at a rate of \(1.1 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches \(95 \mathrm{~km} / \mathrm{h},\) then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at \(-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Assume it stops at each intermediate station for \(22 \mathrm{~s}\).

A robot used in a pharmacy picks up a medicine bottle at \(t=0 .\) It accelerates at \(0.20 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.0 \mathrm{~s},\) then travels without acceleration for \(68 \mathrm{~s}\) and finally decelerates at \(-0.40 \mathrm{~m} / \mathrm{s}^{2}\) for \(2.5 \mathrm{~s}\) to reach the counter where the pharmacist will take the medicine from the robot. From how far away did the robot fetch the medicine?

(III) A bowling ball traveling with constant speed hits the pins at the end of a bowling lane \(16.5 \mathrm{~m}\) long. The bowler hears the sound of the ball hitting the pins \(2.50 \mathrm{~s}\) after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is \(340 \mathrm{~m} / \mathrm{s} ?\)

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