Course: 2.6 统一加速期间流离失所

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统一加速期间流离失所问题
Long runners try to maintain constant with very little or deceleration to conserve .
::长跑者试图保持恒定,

Displacement During Constant Acceleration
::不变加速期间流离失所问题

When acceleration is constant, there are three equations that relate displacement to two of the other three quantities we use to describe motion – time, velocity, and acceleration. These equations only work when acceleration is constant, but there are, fortunately, quite a few cases of motion where the acceleration is constant. One of the most common, if we ignore air , are objects falling due to gravity.
::当加速是恒定的时,有三个方程式与我们用来描述动作的其他3个数量中的2个(时间、速度和加速度)相移。这些方程式只有在加速是恒定的时才起作用,但幸运的是,有不少动作是加速是恒定的。最常见的一个,如果我们忽略空气,就是因重力而坠落的物体。

When an object is moving with constant velocity, the displacement can be found by multiplying the velocity by the time interval, as shown in the equation below
::当一个对象以恒定速度移动时,可以通过将速度乘以时间间隔来找到位移,如下方程式所示

$\begin{aligned} d = v t \end{aligned}$

::d=vt , d=vt , d=vt , d=vt , d=vt , d=vt

If the object is moving with constant acceleration, but not a constant velocity, we can use a derivation of this equation. Instead of using v , as velocity, we must calculate and use the using this equation:
::如果对象在以恒定加速度移动, 但不是以恒定速度移动, 我们可以使用此方程式的引出。与其使用 v, 作为速度, 我们必须计算并使用此方程式 :

$\begin{aligned} v_{ave} = \frac{1}{2} (v_{f} + v_{i}) \end{aligned}$

::vave=12(vf+六)

The distance, then, for uniformly accelerating motion can be found by multiplying the average velocity by the time.
::那么,通过将平均速度乘以时速,可以发现统一加速运动的距离。

$\begin{aligned} d = \frac{1}{2} (v_{f} + v_{i}) (t) (Equation 1) \end{aligned}$

::d=12(vf+vi)(t)(等式1)

We know that the final velocity for constantly accelerated motion can be found by multiplying the acceleration times time and adding the result to the initial velocity, $\begin{aligned} v_{f} = v_{i} + a t \end{aligned}$ .
::我们知道,通过乘以加速时间并将结果添加到初始速度vf=vi+at,可以找到不断加速运动的最后速度。

The second equation that relates displacement, time, initial velocity, and final velocity is generated by substituting this equation into equation 1.
::第二个方程式涉及移位、时间、初始速度和最终速度,第二个方程式是用方程式1替代这一方程式产生的。

Start by distributing the 1/2 in equation 1 through:
::开始将方程1的 1/2 分布到 :

$\begin{aligned} d & = \frac{1}{2} (v_{f} + v_{i}) (t) = \frac{1}{2} v_{f} t + \frac{1}{2} v_{i} t \end{aligned}$

::d=12(vf+vi)(t)=12vft+12vit

We know that $\begin{aligned} v_{f} = v_{i} + a t \end{aligned}$ . Therefore:
::我们知道这个 vf=vi+at。因此:

$\begin{aligned} \frac{1}{2} v_{i} t + \frac{1}{2} (t) (v_{i} + a t) \end{aligned}$
::12vit+12(t)(vi)+at)

$\begin{aligned} = \frac{1}{2} v_{i} t + \frac{1}{2} v_{i} t + \frac{1}{2} a t^{2} \end{aligned}$
::= 12vit+12vit+12at2

$\begin{aligned} = v_{i} t + \frac{1}{2} a t^{2} (Equation 2) \end{aligned}$
::=vit+12at(相等 2)

The third equation is formed by combining $\begin{aligned} v_{f} = v_{i} + a t \end{aligned}$ and $\begin{aligned} d = \frac{1}{2} (v_{f} + v_{i}) (t) \end{aligned}$ . If we solve the first equation for t and then substitute into the second equation, we get
::第三个方程是结合 vf=vi+at 和 d=12 (vf+vi)(t) 组成的。如果我们解决 t 的第一个方程,然后替换为第二个方程,我们就会得到

$\begin{aligned} d = (\frac{1}{2}) (v_{f} + v_{i}) (\frac{v_{f} - v_{i}}{a}) = (\frac{1}{2}) (\frac{v_{f}^{2} - v_{i}^{2}}{a}) \end{aligned}$
::d=( 12) (vf)+vi(vf)-via)=(12) (vf2)-vi2a)

And solving for $\begin{aligned} v {_{f}}^{2} \end{aligned}$ yields
::并解决 vf2 产量

$\begin{aligned} v {_{f}}^{2} = v {_{i}}^{2} + 2 a d (Equation 3) \end{aligned}$

::v2=VI2+2ad(等式 3)

Keep in mind that these three equations are only valid when acceleration is constant. In many cases, the initial velocity can be set to zero and that simplifies the three equations considerably. When acceleration is constant and the initial velocity is zero, the equations can be simplified to:
::请注意, 这三个方程式只有在加速度不变的情况下才有效。在许多情况下, 初始速度可以设定为零, 并且可以大大简化三个方程式。当加速度不变, 初始速度为零时, 这些方程式可以简化为:

$\begin{aligned} d & = \frac{1}{2} v_{f} t \\ d & = \frac{1}{2} a t^{2} and \\ v {_{f}}^{2} & = 2 a d . \end{aligned}$

::d=12 vf td=12 at2和vf2=2ad。

Example 1
::例1

Suppose a planner is designing an airport for small airplanes. Such planes must reach a of 56 m/s before takeoff and can accelerate at 12.0 m/s ² . What is the minimum length for the runway of this airport?
::假设一个规划者正在设计一个小型飞机的机场,这种飞机起飞前必须达到56米/秒,加速速度可达12.0米/秒。这个机场跑道的最低长度是多少?

The acceleration in this problem is constant and the initial velocity of the airplane is zero. Therefore, we can use the equation $\begin{aligned} v {_{f}}^{2} = 2 a d \end{aligned}$ and solve for $\begin{aligned} d \end{aligned}$ .
::这个问题的加速度是恒定的, 飞机的初始速度是零。因此, 我们可以使用公式 vf2=2ad 并解决 d 。

$\begin{aligned} d = \frac{v {_{f}}^{2}}{2 a} = \frac{(56 m/s)^{2}}{(2) (12.0 {m/s}^{2})} = 130 m \end{aligned}$
::d=vf22a=(56米/秒)2(2)(2)-12.0米/秒=130米

Example 2
::例2

How long does it take a car to travel 30.0 m if it accelerates from rest at a rate of 2.00 m/s ² ?
::如果汽车以2.00m/s2的速率从休息速度加速,需要多久才能行驶30.00米/秒?

The acceleration in this problem is constant and the initial velocity is zero, therefore, we can use $\begin{aligned} d = \frac{1}{2} a t^{2} \end{aligned}$ solved for $\begin{aligned} t \end{aligned}$ .
::这个问题的加速度是恒定的, 初始速度为零, 因此,我们可以使用 d=12at2 解决 t 。

$\begin{aligned} t = \sqrt{\frac{2 d}{a}} = \sqrt{\frac{(2) (30.0 m)}{2.00 {m/s}^{2}}} = 5.48 s \end{aligned}$
::t=2da=(2)(3)0.0m)2.00m/s2=5.48s

Example 3
::例3

A baseball pitcher throws a fastball with a speed of 30.0 m/s. Assume the acceleration is uniform and the distance through which the ball is accelerated is 3.50 m. What is the acceleration?
::棒球投手投出速速速为30.0米/秒的速球。假设加速度一致,加速速度的距离为3.50米。加速度是多少?

Since the acceleration is uniform and the initial velocity is zero, we can use $\begin{aligned} v {_{f}}^{2} = 2 a d \end{aligned}$ solve for $\begin{aligned} a \end{aligned}$ .
::由于加速度是统一的, 初始速度为零, 我们可以使用 vf2=2ad 解答。

$\begin{aligned} a = \frac{v_{f}^{2}}{2 d} = \frac{(30.0 m/s)^{2}}{(2) (3.50 m)} = \frac{900. m^{2} / s^{2}}{7.00 m} = 129 {m/s}^{2} \end{aligned}$
::a=vf22d=(30.0m/s)2(2)(2)3.50m)=900.m2/s27.00m=129m/s2

Suppose we plot the velocity versus time graph for an object undergoing . In this first case, we will assume the object started from rest.
::假设我们为正在运行的天体绘制速度和时间图。在第一个情况下, 我们将假设天体从休息开始。

If the object has a uniform acceleration of 6.0 m/s ² and started from rest, then each succeeding second, the velocity will increase by 6.0 m/s. Here is the table of values and the graph.
::如果对象有一个6.0m/s2的统一加速度,并从休息开始,然后每秒,速度将增加6.0m/s。这是数值表和图表。

In displacement versus time graphs, the slope of the line is the velocity of the object. In this case of a velocity versus time graph, the slope of the line is the acceleration. If you take any segment of this line and determine the $\begin{aligned} Δ y \end{aligned}$ to $\begin{aligned} Δ x \end{aligned}$ ratio, you will get 6.0 m/s ² which we know to be the constant acceleration of this object.
::在移位图和时间图中,线的斜度是对象的速度。在速度与时间图的情况下,线的斜度是加速度。如果将线的任何部分取下并确定 y 至 x 的比率,你将得到6.0 m/s2,我们知道这是这个对象的恒定加速度。

We know from geometry that the area of a triangle is calculated by multiplying one-half the base times the height. The area under the curve in the image above is the area of the triangle shown below. The area of this triangle would be calculated by $\begin{aligned} area = (\frac{1}{2}) (6.0 s) (36 m/s) = 108 m \end{aligned}$ .
::我们从几何学中知道,三角形的区域是用基数乘以其高度的一半来计算。以上图像曲线下的区域是以下显示的三角形区域。此三角形的区域将按区域=(12) (6.0 s)(36 m/s)=108米计算。

By going back to equation 2, we know that $\begin{aligned} displacement = \frac{1}{2} a t^{2} \end{aligned}$ .
::通过回到方程2 我们知道迁移=12at2

Using this equation, we can determine that the displacement of this object in the first 6 seconds of travel is $\begin{aligned} = (\frac{1}{2}) (6.0 {m/s}^{2}) (6.0 s)^{2} = 108 m \end{aligned}$ .
::使用这个方程,我们可以确定,在旅行前6秒钟中,该物体的移动位置是=(12) (6.0 m/s2)(6.0 s)2=108 m。

It is not coincidental that this number is the same as the area of the triangle. In fact, the area underneath the curve in a velocity versus time graph is always equal to the displacement that occurs during that time interval.
::此数字与三角形区域相同并非偶然。事实上, 以速度和时间图显示的曲线下区域总是与该时段间隔期间发生的迁移相等。

Use the PLIX Interactive below to analyze the motion of Jane's dog, Sparky, and visualize how acceleration and displacement can be derived from a velocity-time graph:
::用下面的PLIX互动分析Jane的狗Sparky的动作,

Summary
::摘要

There are three equations we can use when acceleration is constant to relate displacement to two of the other three quantities we use to describe motion – time, velocity, and acceleration:

$\begin{aligned} d = (\frac{1}{2}) (v_{f} + v_{i}) (t) \end{aligned}$ (Equation 1)
::d=(12) (vf+vi)(t)(等式1)

$\begin{aligned} d = v_{i} t + \frac{1}{2} a t^{2} \end{aligned}$ (Equation 2)
::d=vit+12at2(等式2)

$\begin{aligned} v {_{f}}^{2} = v {_{i}}^{2} + 2 a d \end{aligned}$ (Equation 3)
::v2=VI2+2ad(等式 3)

::当加速度不变时,我们可以使用三种方程式,将离位与我们用来描述动作-时间、速度和加速度-的另外三种数量中的两种联系起来:d=(12) (vf+vi)(t)(等式1) d=vit+12at2(等式2) vf2=vi2+2ad(等式3)

When the initial velocity of the object is zero, these three equations become:

$\begin{aligned} d = (\frac{1}{2}) (v_{f}) (t) \end{aligned}$ (Equation 1’)
::d=(12) (vf)(t)(等式1)

$\begin{aligned} d = \frac{1}{2} a t^{2} \end{aligned}$ (Equation 2’)
::d=12at2(等式2)

$\begin{aligned} v {_{f}}^{2} = 2 a d \end{aligned}$ (Equation 3’)
::vf2=2ad(等式 3)

::当天体初始速度为零时,这三种方程变为: d=(12) (vf)(t)(等号1) d=12at2(等号2) vf2=2ad(等号3)

The slope of a velocity versus time graph is the acceleration of the object.
::速度与时间图的斜坡是对象的加速度。

The area under the curve of a velocity versus time graph is the displacement that occurs during the given time interval.
::速度与时间图曲线下的区域是特定时间间隔期间发生的位移。

Review
::回顾

An airplane accelerates with a constant rate of 3.0 m/s ² starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the final velocity?
::从21米/秒的速度开始,飞机以3.0米/秒的恒定速率加速。如果加速期间的距离是535米,最后的速度是多少?

An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s ² . What was the velocity of the car when the acceleration first began?
::一辆汽车在484米的距离内用恒定加速度为-8.0米/秒2来休息。当加速度最初开始时,车速是多少?

An airplane starts from rest and accelerates at a constant 3.00 m/s ² for 20.0 s. What is its displacement in this time?
::飞机从休息开始,20秒以3.00m/s2的恒定值加速。

A driver brings a car to a full stop in 2.0 s.

If the car was initially traveling at 22 m/s, what was the acceleration?
::如果汽车最初在22米/秒时行驶,加速度是多少?

How far did the car travel during braking?
::车在刹车时走多远?

::司机将车开到2.0秒的全站。如果汽车最初在22米/秒时行驶,加速度是多少?车在制动时行驶了多远?

Explore More
::探索更多

Use this resource to answer the question that follows.
::使用此资源回答以下问题。

What does the area bounded by a velocity versus time graph represent?
::速度与时间图之间的区域代表什么?

Section outline

General

统一加速期间流离失所问题

Displacement During Constant Acceleration ::不变加速期间流离失所问题

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Displacement During Constant Acceleration
::不变加速期间流离失所问题

Summary
::摘要

Review
::回顾

Explore More
::探索更多