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Chapter 2: Problem 30

A box slides down an incline with uniform acceleration. It starts from rest and attains a speed of \(2.7 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). Find \((a)\) the acceleration and \((b)\) the distance moved in the first \(6.0 \mathrm{~s}\).

Short Answer

Expert verified
Acceleration is 0.9 m/s虏, and distance moved in the first 6.0 s is 16.2 m.

Step by step solution

01

Identify the given values

We are given the initial velocity \( u = 0 \; m/s\) (since the box starts from rest), the final velocity \( v = 2.7 \; m/s \), and the time \( t = 3.0 \; s \).
02

Use the first equation of motion

To find the acceleration, we use the equation \( v = u + at \). Substitute the known values: \( 2.7 = 0 + a \times 3.0 \).
03

Solve for acceleration

Rearrange the equation to find \( a \): \( a = \frac{2.7}{3.0} \). Calculate the acceleration: \( a = 0.9 \; m/s^2 \).
04

Identify time for the second part

For part (b), we need to find the distance moved in \( t = 6.0 \; s \).
05

Use the second equation of motion

The equation for distance is \( s = ut + \frac{1}{2}at^2 \). Since the box starts from rest, \( u = 0 \). Substitute the known values: \( s = 0 \cdot 6.0 + \frac{1}{2} \times 0.9 \times 6.0^2 \).
06

Solve for the distance

Calculate \( s = \frac{1}{2} \times 0.9 \times 36.0 \). This simplifies to \( s = 0.45 \times 36.0 \). Thus, the distance moved is \( s = 16.2 \; m \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to constant acceleration over a period of time. This means the velocity of an object changes by the same amount every second. For example, if a car accelerates uniformly, its speed increases steadily, rather than in fits and starts. In our exercise, the box has a constant or uniform acceleration as it slides down the incline. Some key points about uniform acceleration are:
  • The rate of change of velocity (acceleration) is consistent.
  • Equations of motion under uniform acceleration help predict future position and velocity of an object.
  • Uniform acceleration is often seen in vertical motion under gravity, where the only force is gravity.
Understanding uniform acceleration is vital in solving kinematics problems, where consistently changing speed leads to predictable motion paths.
Equations of Motion
In kinematics, the equations of motion describe the behavior of moving objects. When dealing with uniform acceleration, these equations become particularly useful. There are three main equations used to solve problems:
  • For velocity: \( v = u + at \)This equation shows how final velocity combines initial velocity with acceleration over time.
  • For distance: \( s = ut + \frac{1}{2}at^2 \)It gives the total distance covered, starting from initial velocity and adding the distance due to acceleration.
  • Linking all variables: \( v^2 = u^2 + 2as \)This relates initial and final velocities with displacement, integrating time indirectly.
By using these equations, solving for unknown variables, such as acceleration or distance, becomes much easier.
Initial Velocity
Initial velocity is the velocity of an object at the start of a time period. In the exercise, the box starts from rest on the incline, meaning the initial velocity \( u = 0 \; m/s \). This forms the baseline from which other velocities are measured.Key features of initial velocity in problems:
  • It often sets the stage for interpreting subsequent motion.
  • In many motion problems, if an object starts from rest, initial velocity is zero, simplifying the equations.
  • Changes in velocity over time are heavily influenced by the starting speed of an object.
Knowing the initial velocity helps in applying the equations of motion correctly, especially when calculating resultant speed or displacement changes.
Final Velocity
Final velocity represents the speed of an object at the end of a given time period. In the kinematics problem presented, the box achieves a final velocity \( v = 2.7 \; m/s \) after sliding down for \(3.0 \; s\).Factors to consider about final velocity:
  • It gives insight into how much the object has accelerated over the given time period.
  • Final velocity can help determine total distance traveled when combined with initial velocity and acceleration.
  • In problems with uniform acceleration, final velocity serves as a crucial point to check the consistent application of equations of motion.
Understanding final velocity is key to analyzing the complete motion of an object, offering insights into both the speed and distance covered by an object in motion.

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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 \(15.387\) \(\mathrm{m} .\)

A ball is dropped from rest at a height of \(50 \mathrm{~m}\) above the ground. (a) What is its speed just before it hits the ground? (b) How long does it take to reach the ground? If we can ignore air friction, the ball is uniformly accelerated until it reaches the ground. Its acceleration is downward and is \(9.81\) \(\mathrm{m} / \mathrm{s}^{2}\). Taking down as positive, we have for the trip: $$ y=50.0 \mathrm{~m} \quad a=9.81 \mathrm{~m} / \mathrm{s}^{2} \quad v_{i}=0 $$ (a) \(v_{f y}^{2}=v_{i y}^{2}+2 a y=0+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(50.0 \mathrm{~m})=981 \mathrm{~m}^{2} / \mathrm{s}^{2}\) and so \(v_{f}=31.3 \mathrm{~m} / \mathrm{s}\). (b) From \(a\left(v_{f y}-v_{i y}\right) / t\), $$ t=\frac{v_{f y}-v_{i y}}{a}=\frac{(31.3-0) \mathrm{m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.19 \mathrm{~s} $$ (We could just as well have taken up as positive. How would the calculation have been changed?)

A marble, rolling with speed \(20 \mathrm{~cm} / \mathrm{s}\), rolls off the edge of a table that is \(80 \mathrm{~cm}\) high. (a) How long does it take to drop to the floor? (b) How far, horizontally, from the table edge does the marble strike the floor?

A baseball is thrown straight upward with a speed of \(30 \mathrm{~m} / \mathrm{s}\). (a) How long will it rise? ( \(b\) ) How high will it rise? (c) How long after it leaves the hand will it return to the starting point? ( \(d\) ) When will its speed be \(16 \mathrm{~m} / \mathrm{s}\) ?

A train running along a straight track at \(30 \mathrm{~m} / \mathrm{s}\) is slowed uniformly to a stop in \(44 \mathrm{~s}\). Find the acceleration and the stopping distance.
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