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(I) A car accelerates from \(12 \mathrm{~m} / \mathrm{s}\) to \(21 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~s}\). What was its acceleration? How far did it travel in this time? Assume constant acceleration.

Short Answer

Expert verified
Acceleration: \(1.5 \, \mathrm{m/s^2}\), Distance: \(99 \, \mathrm{m}\).

Step by step solution

01

Identify Known Values

First, we need to identify and list the values given in the problem. The initial velocity \( u \) is \( 12 \, \mathrm{m/s} \), the final velocity \( v \) is \( 21 \, \mathrm{m/s} \), and the time \( t \) is \( 6.0 \, \mathrm{s} \). We have to find the acceleration \( a \) and the distance traveled \( s \).
02

Calculate Acceleration

We use the formula for acceleration given as \( a = \frac{v - u}{t} \), where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time. Substituting the known values, we get \( a = \frac{21 \, \mathrm{m/s} - 12 \, \mathrm{m/s}}{6.0 \, \mathrm{s}} = \frac{9 \, \mathrm{m/s}}{6.0 \, \mathrm{s}} = 1.5 \, \mathrm{m/s^2} \).
03

Calculate Distance Traveled

To find the distance traveled, we use the second equation of motion: \( s = ut + \frac{1}{2}at^2 \). Substituting the known values, we have:\[ s = 12 \, \mathrm{m/s} \times 6.0 \, \mathrm{s} + \frac{1}{2} \times 1.5 \, \mathrm{m/s^2} \times (6.0 \, \mathrm{s})^2 \]Calculating this gives:\[ s = 72 \, \mathrm{m} + \frac{1}{2} \times 1.5 \, \mathrm{m/s^2} \times 36 \, \mathrm{s^2} = 72 \, \mathrm{m} + 27 \, \mathrm{m} = 99 \, \mathrm{m} \].

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Key Concepts

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

Understanding Constant Acceleration
In physics, constant acceleration means that the rate of change of velocity remains uniform over time. This simplifies calculations since the acceleration value does not vary during the motion. Acceleration is denoted by \( a \), and it is usually measured in meters per second squared (\( \mathrm{m/s^2} \)). In the exercise, the car's acceleration is described as constant. This helps us calculate the change in speed and the distance traveled.
In many motion problems, constant acceleration allows the use of specific kinematic equations. These equations make it straightforward to compute unknown variables such as distance or time.
When dealing with motion, constant acceleration can arise in everyday situations such as a car speeding up or slowing down on a straight road. The key takeaway is that when acceleration is constant, calculations become more straightforward, allowing for the efficient use of formulas.
Equations of Motion Explained
Equations of motion are fundamental tools in physics that connect the concepts of velocity, acceleration, time, and displacement. These equations are especially convenient when acceleration is constant, simplifying the study of object movement.
For the car in the exercise, we have two critical equations:
  • Acceleration: \( a = \frac{v-u}{t} \)
  • Distance: \( s = ut + \frac{1}{2}at^2 \)
Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is acceleration, \( t \) is time, and \( s \) is displacement.
By substituting known values into these equations, one can easily find unknown variables, such as acceleration or distance, as demonstrated in the solutions. Learning to use these equations is crucial for solving problems involving linear motion under constant acceleration.
It's essential to identify which variable remains constant and which are changing to apply the right equation effectively.
Initial and Final Velocity Impact
Initial and final velocities are components of motion that describe an object's speed at different times. The initial velocity \( u \) refers to how fast an object is moving at the start of a time interval, while the final velocity \( v \) represents its speed at the end.
In the given exercise, the car starts with an initial velocity of \( 12 \, \mathrm{m/s} \) and reaches a final velocity of \( 21 \, \mathrm{m/s} \). The change from initial to final velocity indicates acceleration, showing how swiftly or gradually the car speeds up.
Knowing both velocities is crucial for computing acceleration using the formula \( a = \frac{v-u}{t} \). It also helps in determining other motion parameters like distance traveled. These factors are intertwined in equations of motion, underlining their importance in kinematic studies.

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Most popular questions from this chapter

In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say \(1.0 \mathrm{~m}\) long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. \(2-48\) ) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) going downhill, and constantly at \(2.8 \mathrm{~m} / \mathrm{s}^{2}\) going uphill. Suppose we have an uphill lie \(7.0 \mathrm{~m}\) from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range \(1.0 \mathrm{~m}\) short to \(1.0 \mathrm{~m}\) long of the cup. Do the same for a downhill lie \(7.0 \mathrm{~m}\) from the cup. What in your results suggests that the downhill putt is more difficult?

(III) Mary and Sally are in a foot race (Fig. \(43 ) .\) When Mary is 22 \(\mathrm{m}\) from the finish line, she has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) and is 5.0 \(\mathrm{m}\) behind Sally, who has a speed of 5.0 \(\mathrm{m} / \mathrm{s}\) . Sally thinksshe has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.50 \(\mathrm{m} / \mathrm{s}^{2}\) to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

(II) A sports car moving at constant speed travels \(110 \mathrm{~m}\) in \(5.0 \mathrm{~s}\). If it then brakes and comes to a stop in \(4.0 \mathrm{~s}\), what is the magnitude of its acceleration in \(\mathrm{m} / \mathrm{s}^{2},\) and in \(g\) 's \(\left(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\right) ?\)

Bill can throw a ball vertically at a speed 1.5 times faster than Joe can. How many times higher will Bill's ball go than Joe's?

(III) The acceleration of a particle is given by \(a=A \sqrt{t}\) where \(A=2.0 \mathrm{~m} / \mathrm{s}^{5 / 2} .\) At \(t=0, v=7.5 \mathrm{~m} / \mathrm{s}\) and \(x=0\) (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, speed and displacement at \(t=5.0 \mathrm{~s} ?\)

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