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Chapter 2: Problem 9

A bus moving in a straight line at a speed of \(20 \mathrm{~m} / \mathrm{s}\) begins to slow at a constant rate of \(3.0 \mathrm{~m} / \mathrm{s}\) each second. Find how far it goes before stopping. Take the direction of motion to be the \(+x\) -direction. For the trip under consideration, \(v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~m} / \mathrm{s}\), \(a=-3.0 \mathrm{~m} / \mathrm{s}^{2}\). Notice that the bus is not speeding up in the positive motion direction. Instead, it is slowing in that direction and so its acceleration is negative (a deceleration). Use $$v_{f x}^{2}=v_{i x}^{2}+2 a x \quad \text { and, hence, } \quad 0=(20 \mathrm{~m} / \mathrm{s})^{2}+2\left(-3.0 \mathrm{~m} / \mathrm{s}^{2}\right) x$$ to find $$x=\frac{-(20 \mathrm{~m} / \mathrm{s})^{2}}{2\left(-3.0 \mathrm{~m} / \mathrm{s}^{2}\right)}=67 \mathrm{~m}$$

Short Answer

Expert verified
The bus travels approximately 67 meters before stopping.

Step by step solution

01

Identify Given Variables

We start by identifying all the given information. Here, the initial velocity \( v_i \) is \( 20 \, \text{m/s} \), the final velocity \( v_f \) is \( 0 \, \text{m/s} \), and the acceleration \( a \) is \( -3.0 \, \text{m/s}^2 \).
02

Choose the Appropriate Equation

We use the kinematic equation \( v_{fx}^2 = v_{ix}^2 + 2ax \) to find the distance \( x \). This equation relates velocity, acceleration, and displacement, which are all variables in our problem.
03

Substitute Known Values

Substitute the known values into the equation: \( 0 = (20 \, \text{m/s})^2 + 2(-3.0 \, \text{m/s}^2)x \). The final velocity is \( 0 \, \text{m/s} \), the initial velocity is \( 20 \, \text{m/s} \), and the acceleration is \( -3.0 \, \text{m/s}^2 \).
04

Rearrange the Equation and Solve for \( x \)

Solving the equation for \( x \), we rearrange it to find: \[ x = \frac{-(20 \, \text{m/s})^2}{2 \times (-3.0 \, \text{m/s}^2)} \].
05

Calculate the Distance

Calculate the value using the formula: \[ x = \frac{-400}{-6} = 66.67 \approx 67 \, \text{m} \]. Therefore, the bus travels approximately 67 meters before stopping.

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Key Concepts

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

Deceleration
When an object, such as a bus, reduces its speed over time, it is undergoing deceleration. This is also often termed as negative acceleration. Unlike acceleration, where the object's speed increases, deceleration means the object is slowing down. In our example, the bus is decreasing its speed at a constant rate of
  • **Rate:** The deceleration rate is \[ -3.0 \, \text{m/s}^2 \].This negative sign indicates that the force acting on the bus is in the opposite direction of its motion.
Understanding deceleration is critical for predicting how long it will take for an object to stop completely and how far it will travel during this time. Recognizing that deceleration is a constant rate helps simplify calculations using kinematic equations.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the forces causing this motion. These equations are essential for solving problems involving constant acceleration, including deceleration. In this exercise, we dealt with the equation:
  • \[ v_{fx}^2 = v_{ix}^2 + 2ax \]
This specific equation links
  • initial velocity \( v_{ix} \),
  • final velocity \( v_{fx} \),
  • displacement \( x \), and
  • acceleration \( a \).
Using kinematic equations, we can determine unknowns -- in this case, the distance the bus travels before stopping. It is vital to correctly substitute known values into the equation and rearrange it to solve for the unknown variable. Equations like this provide a straightforward way to relate critical variables of motion such as speed, stopping distance, and time.
Initial and Final Velocities
Understanding initial and final velocities is a core part of analyzing motion. Initial velocity is the speed at which an object starts its observation. Here, the bus has an initial velocity of
  • \( v_i = 20 \, \text{m/s} \).
Final velocity is the speed when the object has reached the end of the observation phase. For the bus, it comes to a complete stop, so the final velocity is
  • \( v_f = 0 \, \text{m/s} \).
Knowing both velocities allows us to compute the change in velocity, which, combined with acceleration, helps to find the displacement over this period. Initial and final velocities are fundamental for calculating how long it takes and how far an object travels under specific acceleration conditions.

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Most popular questions from this chapter

A skier starts from rest and slides down a mountain side along a straight descending path \(9.0 \mathrm{~m}\) long in \(3.0 \mathrm{~s}\). In what time after starting will the skier acquire a speed of \(24 \mathrm{~m} / \mathrm{s}\) ? Assume that the acceleration is constant and the entire run is straight and at a fixed incline. We must find the skier's acceleration from the data concerning the \(3.0 \mathrm{~s}\) trip. Taking the direction of motion down the inclined path as the \(+x\) -direction, we have \(t=3.0 \mathrm{~s}, v_{i x}=0\), and \(x=9.0 \mathrm{~m} .\) Then \(x=v_{i x} t+\frac{1}{2} a t^{2}\) gives $$a=\frac{2 x}{t^{2}}=\frac{18 \mathrm{~m}}{(3.0 \mathrm{~s})^{2}}=2.0 \mathrm{~m} / \mathrm{s}^{2}$$ We can now use this value of \(a\) for the longer trip, from the starting point to the place where \(v_{f x}=24 \mathrm{~m} / \mathrm{s}\). For this trip, \(v_{i x}=0, v_{f x}=24 \mathrm{~m} / \mathrm{s}, a=2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Then, from \(v_{f}=v_{i}+a t\), $$ t=\frac{v_{f x}-v_{i x}}{a}=\frac{24 \mathrm{~m} / \mathrm{s}}{2.0 \mathrm{~m} / \mathrm{s}^{2}}=12 \mathrm{~s}. $$

A car is traveling at \(20.0 \mathrm{~m} / \mathrm{s}\) when the driver slams on the brakes and brings it to a straight-line stop in \(4.2 \mathrm{~s}\). What is the magnitude of its average acceleration? The defining scalar equation is \(a_{a v}=\left(v_{f}-v_{i}\right) / t\). Note that the final speed is zero. Here the initial speed is greater than the final speed, so we can expect the acceleration to be negative: $$a_{a v}=\frac{0.0 \mathrm{~m} / \mathrm{s}-20.0 \mathrm{~m} / \mathrm{s}}{4.2 \mathrm{~s}}=-4.76 \mathrm{~m} / \mathrm{s}^{2}$$ Because the time is provided with only two significant figures, the answer is \(-4.8 \mathrm{~m} / \mathrm{s}^{2}\).

A stone is thrown straight upward and it rises to a maximum height of \(20 \mathrm{~m}\). With what speed was it thrown? Take \(u p\) as the positive \(y\) -direction. The stone's velocity is zero at the top of its path. Then \(v_{f y}=0, y=20 \mathrm{~m}\), \(a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\). (The minus sign arises because the acceleration due to gravity is always downward and we have taken \(u p\) to be positive.) Use \(v_{f y}^{2}=v_{i y}^{2}+2 a y\) to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$

A stone is thrown straight upward with a speed of \(20 \mathrm{~m} / \mathrm{s}\). It is caught on its way down a 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 \(u p\) 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 $$\begin{array}{l} v_{f y}^{2}=(20 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~m})=302 \mathrm{~m}^{2} / \mathrm{s}^{2} \\\ v_{f y}=\pm \sqrt{302 \mathrm{~m}^{2} / \mathrm{s}^{2}}=-17 \mathrm{~m} / \mathrm{s} \end{array}$$ 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 \(v_{f y}3.\)

A hose lying on the ground shoots a stream of water upward at an angle of \(40^{\circ}\) to the horizontal. The speed of the water is \(20 \mathrm{~m} / \mathrm{s}\) as it leaves the hose. How high up will it strike a wall that is a horizontal distance of \(8.0 \mathrm{~m}\) away?
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