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A car accelerates from rest at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.0 \mathrm{~s}\). (a) What is the speed of the car at the end of that time? (b) How far does the car travel in this time?

Short Answer

Expert verified
(a) 10.0 m/s, (b) 25.0 m.

Step by step solution

01

Identify given information

We know the car starts from rest, so the initial velocity, \( u = 0 \). The constant acceleration is given as \( a = 2.0 \, \text{m/s}^2 \), and the time over which the car accelerates is \( t = 5.0 \, \text{s} \).
02

Calculate final speed (Part a)

To find the final speed of the car, use the equation for velocity, \( v = u + at \). Substituting the known values: \( v = 0 + (2.0 \, \text{m/s}^2)(5.0 \, \text{s}) = 10.0 \, \text{m/s} \). So, the final speed of the car is \( 10.0 \, \text{m/s} \).
03

Calculate distance travelled (Part b)

To find the distance travelled, use the second equation of motion: \( s = ut + \frac{1}{2}at^2 \). Substituting the given values, \( s = 0 \cdot 5.0 + \frac{1}{2}(2.0)(5.0)^2 = \frac{1}{2}(2.0)(25.0) = 25.0 \, \text{m} \). Hence, the car travels a distance of \( 25.0 \, \text{m} \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration, also known as constant acceleration, is a crucial topic in kinematics. This is when an object's speed increases by the same amount each second.
When dealing with uniform acceleration, three main types of motions can occur:
  • Acceleration: Increasing speed over time.
  • Deceleration: Decreasing speed over time.
  • Constant Speed: No change in speed over time.
In our example with the car, the acceleration is uniform because its speed changes at a constant rate of \(2.0 \, \text{m/s}^2\).
Whenever you hear "uniform acceleration," think of a straight line graph of speed versus time, showing a constant slope. Understanding this concept helps in predicting how fast an object will move over a certain period, which is essential in solving problems involving equations of motion.
Equations of Motion
The equations of motion are a set of formulas used to calculate different aspects of moving objects under uniform acceleration. These are fundamental for solving problems in kinematics.
The three primary equations are:
  • \(v = u + at\)
  • \(s = ut + \frac{1}{2}at^2\)
  • \(v^2 = u^2 + 2as\)
Let's see how these work:- The first equation calculates the final velocity of the object.- The second equation helps to determine the distance traveled by the object.- The third equation provides a means to relate final velocity with distance without involving time.
In our car example, the first two equations were directly used to find the final velocity and distance.
Grasping these formulas is key because once mastered, they allow you to tackle a wide array of physics problems involving uniform acceleration.
Initial Velocity
Initial velocity is the velocity of an object before it begins to undergo any change caused by forces like acceleration. It is often denoted by \(u\) in equations of motion.
In many scenarios, such as with the car in our example, the object starts from rest. This means the initial velocity \(u\) is \(0 \text{ m/s}\).
Initial velocity can greatly influence how an object behaves over time under acceleration:
  • If the initial velocity is zero, it simplifies calculations as seen in the car's equation of motion.
  • When the initial velocity is not zero, it must be accounted for in calculations to correctly determine the final velocity or position.
Understanding initial velocity is crucial as it forms the basis from which other calculations, like those for final velocity or displacement, are made.
Final Velocity
Final velocity is the velocity an object reaches after undergoing acceleration over a period of time. It's what you calculate using the first equation of motion: \(v = u + at\).
In cases where an object starts from rest, like in our example with the car:
  • The initial velocity \(u\) is \(0 \mathrm{~m/s}\), making it easier to compute the final velocity.
By plugging the known acceleration and time into this equation:- The car ends up with a velocity of \(10.0 \mathrm{~m/s}\) after \(5 \text{ seconds}\).
Final velocity is significant since it helps predict how an object will continue to move, such as understanding if it will stop, slow down, or speed up over the next interval of time. Understanding this concept is essential for analyzing motion comprehensively.

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Most popular questions from this chapter

A test rocket containing a probe to determine the composition of the upper atmosphere is fired vertically upward from an initial position at ground level. During the time \(t\) while its fuel lasts, the rocket ascends with a constant upward acceleration of magnitude \(2 g .\) Assume that the rocket travels to a small enough height that the Earth's gravitational force can be considered constant. (a) What are the speed and height, in terms of \(g\) and \(t,\) when the rocket's fuel runs out? (b) What is the maximum height the rocket reaches in terms of \(g\) and \(t ?(\mathrm{c})\) If \(t=30.0 \mathrm{~s},\) calculate the rocket's maximum height.

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A senior citizen walks \(0.30 \mathrm{~km}\) in \(10 \mathrm{~min}\), going around a shopping mall. (a) What is her average speed in meters per second? (b) If she wants to increase her average speed by \(20 \%\) when walking a second lap, what would her travel time in minutes have to be?

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