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Chapter 2: Problem 35

An object moving at \(13 \mathrm{~m} / \mathrm{s}\) slows uniformly at the rate of \(2.0 \mathrm{~m} / \mathrm{s}\) each second for a time of \(6.0\) s. Determine \((a)\) its final speed, \((b)\) its average speed during the \(6.0 \mathrm{~s}\), and \((c)\) the distance moved in the \(6.0 \mathrm{~s}\).

Short Answer

Expert verified
(a) 1 m/s, (b) 7 m/s, (c) 42 m

Step by step solution

01

Calculate Final Speed

To find the final speed of the object, we use the formula for uniformly accelerated (or decelerated, in this case) motion: \[ v_f = v_i + a \cdot t \]where:- \( v_f \) is the final velocity- \( v_i = 13 \text{ m/s} \) is the initial velocity- \( a = -2.0 \text{ m/s}^2 \) (deceleration is negative)- \( t = 6.0 \text{ s} \) is the timeSubstitute the values into the equation:\[ v_f = 13 + (-2.0) \times 6 = 13 - 12 = 1 \text{ m/s} \]
02

Calculate Average Speed

The average speed of the object is calculated by taking the mean of the initial and final speeds, because the acceleration is constant. The formula is:\[ v_{\text{avg}} = \frac{v_i + v_f}{2} \]Substitute the known values:\[ v_{\text{avg}} = \frac{13 + 1}{2} = \frac{14}{2} = 7 \text{ m/s} \]
03

Calculate Distance Moved

To find the total distance moved, we use the formula:\[ d = v_{\text{avg}} \times t \]where:- \( v_{\text{avg}} = 7 \text{ m/s} \) (calculated from Step 2)- \( t = 6.0 \text{ s} \)Plug in the numbers:\[ d = 7 \times 6 = 42 \text{ m} \]

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools in physics, especially when dealing with motion problems that show uniform acceleration. They help us describe the motion of objects and predict future movements based on current information.
These equations relate four important motion variables: initial velocity (\( v_i \)), final velocity (\( v_f \)), time (\( t \)), acceleration (\( a \)), and displacement or distance (\( d \)).
Here's a quick look at the main kinematic equations:
  • Final velocity: \[ v_f = v_i + a \cdot t \]
  • Displacement: \[ d = v_i \cdot t + 0.5 \cdot a \cdot t^2 \]
  • Average velocity: \[ v_{\text{avg}} = \frac{v_i + v_f}{2} \]
  • Another form for displacement: \[ d = v_{\text{avg}} \cdot t \]
These equations allow us to solve problems when certain conditions are known, like initial speed, time, and acceleration, enabling calculations of unknown values such as final velocity or distance.
Final Velocity
Final velocity (\( v_f \)) is a vital concept in kinematics, depicting the speed of an object at a specific time after it has been accelerated or decelerated for a certain period.
To compute the final velocity in situations of uniform acceleration, the kinematic equation \( v_f = v_i + a \cdot t \) is used.
Here:
  • \( v_i \), the initial velocity, is the speed at which the object starts. In our example: 13 m/s.
  • \( a \), the acceleration, tells us how much the speed changes over time. Deceleration means \( a \) is negative. Here: -2.0 m/s虏.
  • \( t \), time, refers to how long the acceleration or deceleration process takes, like 6 seconds here.
By substituting these known values into the formula, we find that the final velocity of the object is 1 m/s, indicating that it has slowed down from its initial speed.
Average Speed
Average speed is a measure of the total distance traveled divided by the total time taken. It gives us an insight into how fast an object is moving on average over a specific duration.
When acceleration is uniform, finding average speed is straightforward as it's simply the mean of the initial and final velocities:\( v_{\text{avg}} = \frac{v_i + v_f}{2} \)
For our object:
  • The initial velocity, \( v_i \), is 13 m/s.
  • The final velocity, \( v_f \), after deceleration, is 1 m/s.
Putting these into the formula gives us:\[ v_{\text{avg}} = \frac{13 + 1}{2} = 7 \text{ m/s} \]This average speed tells us that, over the 6 seconds of movement, the object traveled at an average speed of 7 meters every second.
Distance Traveled
Distance traveled, or displacement, is a fundamental aspect of motion found by determining how far an object has moved over time. We often compute this by using both the average speed calculation and the time over which motion occurs.
  • The distance, \( d \), for an object moving with uniform acceleration is often calculated as:\[ d = v_{\text{avg}} \cdot t \]
  • The average speed, \( v_{\text{avg}} \), in our example is 7 m/s.
  • The time of travel, \( t \), is 6 seconds.
Substitute these into the formula:\[ d = 7 \times 6 = 42 \text{ m} \]This result illustrates that over the 6-second period, the object moved a total distance of 42 meters. Understanding the distance formula enhances our comprehension of how speed and time interconnect to influence movement outcomes.

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Most popular questions from this chapter

A stone is thrown straight upward with a speed of \(20 \mathrm{~m} / \mathrm{s}\). It is caught on its way down at a point \(5.0 \mathrm{~m}\) above where it was thrown. (a) How fast was it going when it was caught? ( \(b\) ) How long did the trip take? The situation is shown in Fig. 2-3. Take up as positive. Then, for the trip that lasts from the instant after throwing to the instant before catching, \(v_{i y}=20 \mathrm{~m} / \mathrm{s}, y=+5.0 \mathrm{~m}\) (since it is an upward displacement), \(a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\) (a) Use \(v_{f y}^{2}=v_{i y}^{2}+2 a y\) to compute $$ \begin{array}{l} v_{f j}^{2}=(20 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~m})=302 \mathrm{~m}^{2} / \mathrm{s}^{2} \\\ v_{f j}=\pm \sqrt{302 \mathrm{~m}^{2} / \mathrm{s}^{2}}=-17 \mathrm{~m} / \mathrm{s} \end{array} $$ Take the negative sign because the stone is moving downward, in the negative direction, at the final instant. (b) To find the time, use \(a=\left(v_{f y}-v_{i y}\right) / t\) and so $$ t=\frac{(-17.4-20) \mathrm{m} / \mathrm{s}}{-9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.8 \mathrm{~s} $$ Notice that we retain the minus sign on \(v_{f y}\). You can check your work by dividing the problem into two parts, the trip up to peak altitude and the trip down from peak. The peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 g=-(20\) \(\mathrm{m} / \mathrm{s})^{2} / 2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=20.38736 \mathrm{~m}\). [Hint: Don't round off to two figures mid-calculation.] Now drop the stone from \(y_{p}\) so it falls a distance \((20.38736 \mathrm{~m})-(5.0 \mathrm{~m})=15.38736 \mathrm{~m}\), at which point it will be moving- - from Eq. (2.6) with down as plus-at \(v_{f}^{2}=2 g s=\) \(2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(15.38736 \mathrm{~m})=301.900 \mathrm{~m}^{2} / \mathrm{s}^{2}\), and so \(v_{f}=17.4 \mathrm{~m} / \mathrm{s}=\) \(17 \mathrm{~m} / \mathrm{s}\). Similarly, you can calculate the total time of flight, which equals the time to reach peak altitude plus the time to fall \(15.387\) \(\mathrm{m} .\)

The speed of a train is reduced uniformly from \(15 \mathrm{~m} / \mathrm{s}\) to \(7.0 \mathrm{~m} / \mathrm{s}\) while traveling a distance of \(90 \mathrm{~m} .(a)\) Compute the acceleration. (b) How much farther will the train travel before coming to rest, provided the acceleration remains constant? Take the direction of motion to be the \(+x\) -direction. (a) We have \(v_{i x}=15 \mathrm{~m} / \mathrm{s}, v_{f x}=7.0 \mathrm{~m} / \mathrm{s}, x=90 \mathrm{~m}\). Then \(v^{2} f_{x}=v^{2} i x\) \(+2 a x\) gives \(a=-0.98 \mathrm{~m} / \mathrm{s}^{2}\) (b) The new conditions \(v_{i x}=7.0 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-0.98 \mathrm{~m} / \mathrm{s}^{2}\) now obtain. Then leads to

A stone is thrown straight downward with initial speed \(8.0 \mathrm{~m} / \mathrm{s}\) from a height of \(25 \mathrm{~m}\). Find \((a)\) the time it takes to reach the ground and \((b)\) the speed with which it strikes.

A ball is thrown upward at an angle of \(30^{\circ}\) to the horizontal and lands on the top edge of a building that is \(20 \mathrm{~m}\) away. The top edge is \(5.0 \mathrm{~m}\) above the throwing point. How fast was the ball thrown?

A ball that is thrown vertically upward on the Moon returns to its starting point in \(4.0 \mathrm{~s}\). The acceleration due to gravity there is \(1.60\) \(\mathrm{m} / \mathrm{s}^{2}\) downward. Find the ball's original speed. Take up as positive. For the trip from beginning to end, \(y=0\) (it ends at the same level it started at), \(a=-1.60 \mathrm{~m} / \mathrm{s}^{2}, t=4.0 \mathrm{~s}\). Use \(y=v_{i y} t+\frac{1}{2} a t^{2}\) to find $$ 0=v_{i y}(4.0 \mathrm{~s})+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~s})^{2} $$ from which \(v_{i y}=3.2 \mathrm{~m} / \mathrm{s}\).
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