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Chapter 2: Problem 34

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.

Short Answer

Expert verified
Acceleration is -0.682 m/s虏, and stopping distance is 659.78 m.

Step by step solution

01

Identify What is Given

We are given the initial velocity of the train (\(v_i = 30 \, \text{m/s}\)), the final velocity (\(v_f = 0 \, \text{m/s}\)), and the time taken to stop (\(t = 44 \, \text{s}\)). Our task is to find the acceleration and the stopping distance.
02

Use the Formula for Acceleration

Since the train slows uniformly, we can use the formula \(a = \frac{v_f - v_i}{t}\). Plugging in the given values, we get:\[ a = \frac{0 - 30}{44} = \frac{-30}{44} \approx -0.6818 \, \text{m/s}^2.\]This means the train has an acceleration of approximately -0.682 m/s虏.
03

Apply the Equation for Stopping Distance

We want to find the distance (\(d\)) the train travels as it comes to a stop. We'll use the equation: \[d = v_i t + \frac{1}{2} a t^2.\] Substituting known values in, we have:\[ d = 30 \times 44 + \frac{1}{2} \times (-0.6818) \times (44)^2.\]
04

Calculate the Stopping Distance

Continuing with the calculation:\[ d = 1320 + \frac{1}{2} \times (-0.6818) \times 1936.\]This becomes:\[ 1320 - \frac{1}{2} \times 1320.44 \approx 1320 - 660.22 = 659.78 \, \text{m}.\]Thus, the stopping distance of the train is approximately 659.78 meters.

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Key Concepts

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

Understanding Kinematics
Kinematics is the study of the motion of objects without considering the forces that cause the motion. It deals with concepts like displacement, velocity, and acceleration to describe how objects move.
Kinematics problems often involve determining how various factors change over time. In our exercise, the train鈥檚 motion is a kinematics problem because it focuses on how the speed and position of the train change as time progresses.
In this specific scenario, we utilize kinematic equations to solve for parameters like final velocity, acceleration, and stopping distance. The key is to relate these values through the use of formulas that describe uniform motion, such as the basic equations of motion under constant acceleration.
Diving into Velocity
Velocity is a vector quantity that describes the speed of an object in a specific direction. It's crucial in kinematics because it defines how fast something is moving and where it's headed.
In the train problem, the initial velocity (\(v_i\) is given as 30 m/s, which indicates the train's speed before it begins to slow down. The final velocity (\(v_f\)), however, is 0 m/s, as the train comes to a complete stop.
The change in velocity is a critical component of calculating acceleration, as it tells us how much the speed varies over a period of time. In this case, knowing both initial and final velocities helps compute the uniform acceleration of the train as it races to a halt.
Decoding Acceleration
Acceleration is the rate at which an object's velocity changes with time. It can be positive (speeding up) or negative (slowing down), often referred to as deceleration.
The formula for acceleration is \(a = \frac{v_f - v_i}{t}\). For the train example, the initial velocity is 30 m/s, the final velocity is 0 m/s, and the time taken is 44 seconds. Plugging these values into the formula gives us an acceleration of approximately -0.682 m/s虏.
The negative sign indicates that the train is slowing down, as the velocity decreases over time. Understanding acceleration is crucial for solving other aspects of the problem, including calculating how far an object moves while decelerating.
Exploring Stopping Distance
Stopping distance is the total distance an object travels from the point where it begins to decelerate until it comes to a complete stop.
To calculate the stopping distance for the train, the equation used is \(d = v_i t + \frac{1}{2} a t^2\). This formula considers both the initial velocity and the acceleration over the given time period.
With the train's initial velocity at 30 m/s, time as 44 seconds, and acceleration as -0.6818 m/s虏, we compute stopping distance as approximately 659.78 meters.
Stopping distance provides insight into how far an object will move before it halts, taking into account its initial speed and the rate of deceleration. Understanding this concept helps in planning real-world scenarios, ensuring safety and efficiency.

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Most popular questions from this chapter

A truck's speed increases uniformly from \(15 \mathrm{~km} / \mathrm{h}\) to \(60 \mathrm{~km} / \mathrm{h}\) in 20 s. Determine \((a)\) the average speed, \((b)\) the acceleration, and \((c)\) the distance traveled, all in units of meters and seconds. For the 20-s trip under discussion, taking \(+x\) to be in the direction of motion, $$ \begin{array}{l} v_{i x}=\left(15 \frac{\mathrm{k} m}{\not \mathrm{h}}\right)\left(1000 \frac{\mathrm{m}}{\mathrm{k} \not \mathrm{\gamma}}\right)\left(\frac{1}{3600} \frac{\not \mathrm{K}}{\mathrm{s}}\right)=4.17 \mathrm{~m} / \mathrm{s} \\ v_{f x}=60 \mathrm{~km} / \mathrm{h}=16.7 \mathrm{~m} / \mathrm{s} \end{array} $$ (a) \(v_{a v}=\frac{1}{2}\left(v_{i x}+v_{f x}\right)=\frac{1}{2}(4.17+16.7) \mathrm{m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}\) (b) \(\quad a=\frac{v_{f x}-v_{i x}}{t}=\frac{(16.7-4.17) \mathrm{m} / \mathrm{s}}{20 \mathrm{~s}}=0.63 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(\quad x=v_{\text {av }} t=(10.4 \mathrm{~m} / \mathrm{s})(20 \mathrm{~s})=208 \mathrm{~m}=0.21 \mathrm{~km}\)

A plane starts from rest and accelerates uniformly in a straight line along the ground before takeoff. It moves \(600 \mathrm{~m}\) in \(12 \mathrm{~s}\). Find \((a)\) the acceleration, \((b)\) speed at the end of \(12 \mathrm{~s}\), and \((c)\) the distance moved during the twelfth second.

A baseball is thrown straight upward on the Moon with an initial speed of \(35 \mathrm{~m} / \mathrm{s}\). Compute \((a)\) the maximum height reached by the ball, (b) the time taken to reach that height, \((c)\) its velocity \(30 \mathrm{~s}\) after it is thrown, and \((d)\) when the ball's height is \(100 \mathrm{~m}\). Take up as positive. At the highest point, the ball's velocity is zero. (a) From \(v_{f y}^{2}=v_{i y}^{2}+2 a y\), since \(g=1.60 \mathrm{~m} / \mathrm{s}^{2}\) on the Moon, $$ 0=(35 \mathrm{~m} / \mathrm{s})^{2}+2\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=0.38 \mathrm{~km} $$ (b) From \(v_{f y}=v_{i y}+a t\) $$ 0=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t \quad \text { or } \quad t=22 \mathrm{~s} $$ (c) From \(v_{f y}=v_{i y}+\) at $$ v_{f j}=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(30 \mathrm{~s}) \quad \text { or } \quad v_{f y}=-13 \mathrm{~m} / \mathrm{s} $$ Because \(v_{f}\) is negative and we are taking up as positive, the velocity is directed downward. The ball is on its way down at \(t\) \(=30 \mathrm{~s} .\) (d) From \(y=v_{i y} t+\frac{1}{2} a t^{2}\) we have $$ 100 \mathrm{~m}=(34 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \quad \text { or } \quad 0.80 t^{2}-35 t+100=0 $$ By use of the quadratic formula, Or $$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ t &=\frac{35 \pm \sqrt{35^{2}-4(0.80) 100}}{2(0.80)}=\frac{35 \pm 30.08}{1.60} \end{aligned} $$ we find \(t=3.1 \mathrm{~s}\) and \(41 \mathrm{~s}\). At \(t=3.1 \mathrm{~s}\) the ball is at \(100 \mathrm{~m}\) and ascending; at \(t=41 \mathrm{~s}\) it is at the same height but descending.

If a vehicle accelerates at \(10.0 \mathrm{~m} / \mathrm{s}^{2}\) from rest for \(20.0 \mathrm{~s}\), how far will it travel in the process? [Hint: You are given \(a, u_{i}\), and \(t\), and you need to find s.]

(a) Find the range \(x\) of a gun that fires a shell with muzzle velocity \(u\) at an angle of elevation \(\theta\). ( \(b\) ) Find the angle of elevation \(\theta\) of a gun that fires a shell with a muzzle velocity of \(120 \mathrm{~m} / \mathrm{s}\) and hits a target on the same level but \(1300 \mathrm{~m}\) distant. (See \(\underline{\text { Fig. } 2-8 .}\) ) (a) Let \(t\) be the time it takes the shell to hit the target. Then, \(x=v_{i x}\) tor \(t=x / v_{i x}\). Consider the vertical motion alone, and take \(u p\) as positive. When the shell strikes the target, $$ \text { Vertical displacement }=0=v_{i y} t+\frac{1}{2}(-g) t^{2} $$ Solving this equation gives \(t=2 v_{i y} / g .\) But \(t=x / v_{i x}\), so $$ \frac{x}{v_{i x}}=\frac{2 v_{i y}}{g} \quad \text { or } \quad x=\frac{2 v_{i x} v_{i y}}{g}=\frac{2\left(v_{i} \cos \theta\right)\left(v_{i} \sin \theta\right)}{g} $$ wherein \(g\) is positive. The formula \(2 \sin \theta \cos \theta=\sin 2 \theta\) can be used to simplify this. After substitution, $$ x=\frac{v_{i}^{2} \sin 2 \theta}{g} \text { . } $$ The maximum range corresponds to \(\theta=45^{\circ}\), since \(\sin 2 \theta\) has a maximum value of 1 when \(2 \theta=90^{\circ}\) or \(\theta=45^{\circ}\). (b) From the range equation found in \((a)\), $$ \sin 2 \theta=\frac{g x}{v_{i}^{2}}=\frac{\left(9.18 \mathrm{~m} / \mathrm{s}^{2}\right)(1300 \mathrm{~m})}{(120 \mathrm{~m} / \mathrm{s})^{2}}=0.886 $$ Therefore, \(2 \theta=\arcsin 0.886=62^{\circ}\) and so \(\theta=31^{\circ}\).
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