2.5 统一加速式统一加速式

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  统一加速

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  After jumping, this cliff diver undergoes effective free fall. Cliff diving is a sport in which athletes twist and flip on their way down. Even with the air , these divers will accelerate the whole way down.
  ::跳跃后,这个悬崖潜水员自由坠落。克里夫潜水是一种运动,运动员们在运动中摇晃和翻转。即使有空气,这些潜水员也会加速整个下降速度。

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  Uniform Acceleration
  ::统一加速

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  that does not change in time is called uniform or constant acceleration.   The at the beginning of the time interval is called initial velocity, $\begin{array}{r}{v}_i\end{array}$ , and the velocity at the end of the time interval is called final velocity, $\begin{array}{r}{v}_f\end{array}$ .  In a velocity versus time graph for uniform acceleration, the slope of the line is the acceleration. The equation that describes the curve is  $\begin{array}{r}{v}_f=v_i+at\end{array}$ .
  ::不改变时间的,称为统一加速度或恒定加速度。时间间隔开始时称为初始速度,vi,时间间隔结束时的速度称为最终速度, vf。在速度和时间图中,对于统一加速度,线的斜度是加速度。描述曲线的方程式是 vf=vi+at。

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  Example 1
  ::例1

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  If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s ² for 2.5 s, what is the final velocity?
  ::如果速度为4.0米/秒的汽车以2.5秒4.0米/秒的速度加速加速,最后的速度是多少?

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  Solution:
  ::解决方案 :

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  $\begin{array}{r}{v}_f=v_i+at=4.0\text{m/s}+(4.0{\text{m/s}}^2)(2.5\text{s})=4.0\text{m/s}+10.\text{m/s}=14\text{m/s}\end{array}$
  ::vf=vi+at=4.0m/s+(4.0m/s2)(2.5ss)=4.0m/s+10.m/s=14m/s)

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  Example 2 
  ::例2

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  A car accelerates with uniform acceleration from 11.1 m/s to 22.2 m/s in 5.0 s. (a) What was the acceleration and (b) how far did it travel during the acceleration?
  ::5.0秒(a) 加速度是多少,(b) 加速度是多少?

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  Solution: 
  ::解决方案 :

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  (a)  $\begin{array}{r}a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{22.2\text{m/s}-11.1\text{m/s}}{5.0\text{s}}=2.2{\text{m/s}}^2\end{array}$
  :a) avt=22.2 m/s-11.1 m/s5.0 s=2.2 m/s2

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  (b)  We can find the traveled by  $\begin{array}{r}d=v_{i}t+\frac{1}{2}a{t}^2\end{array}$ , or we could find the distance traveled by determining the and multiply it by the time.
  :b) 我们能够找到 d=vit+12at2 所穿越的距离,或者我们可以通过在时间上确定和乘以所穿越的距离。

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  $$\begin{array}{rl}d& =v_{i}t+\frac{1}{2}a{t}^2\\ & =(11.1\text{m/s})(5.0\text{s})+\left(\frac{1}{2}\right)(2.2{\text{m/s}}^2)(5.0\text{s}{)}^2\\ & =55.5\text{m}+27.8\text{m}\\ & =83\text{m}\end{array}$$

  
  ::d=vit+12at2=(11.1米/秒)(5.0秒)+(12)(2.2米/秒)(5.0秒)2=55.5米+27.8米=83米)

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  $\begin{array}{r}d=(v_{\text{avg}})(t)=(16.6\text{m/s})(5.0\text{s})=83\text{m}\end{array}$
  ::d=(vavg)(t)=(16.6 m/s)(5.0 s)=83 m

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  Example 3
  ::例3

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  If a cart slows from 22.0 m/s with an acceleration of -2.0 m/s ² , how long does it require to get to 4 m/s?
  ::如果一个推车慢速22.0m/s,加速度为-2.0m/s2, 加速度达到4m/s需要多长时间?

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  Solution:
  ::解决方案 :

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  $\begin{array}{r}t=\frac{v_f-v_i}{a}=\frac{-18\text{m/s}}{-2.0{\text{m/s}}^2}=9.0\text{s}\end{array}$
  ::t=vf-via=18 m/s-2.0 m/s2=9.0 s

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  Use the simulation below to further explore the uniform acceleration of a cliff diver. Observe the relationship between position, velocity, and acceleration as the diver falls under the influence of gravity. Begin by turning off the air resistance and see what happens when the diver falls as if he is in a vacuum. Then, if you are up for a challenge, turn the air resistance on and consider its effects on the diver.
  ::使用下面的模拟来进一步探索悬崖潜水员的统一加速度。 当潜水员处于重力影响之下时, 观察位置、 速度和加速度之间的关系 。 开始关闭空气阻力, 并观察潜水员跌倒时会发生什么情况 。 然后, 如果您要面对挑战, 请打开空气阻力, 并考虑其对潜水员的影响 。

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  Summary
  ::摘要

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  • Acceleration that does not change in time is uniform, or constant, acceleration. 
    ::不改变时间的加速是统一或恒定的加速。
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  • The equation relating initial velocity, final velocity, time, and acceleration is $\begin{array}{r}{v}_f=v_i+at\end{array}$ .
    ::与初始速度、最终速度、时间和加速度有关的方程式是 vf=vi+at。

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  Review
  ::回顾

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  1. If an object has zero acceleration, does that mean it has zero velocity?  Give an example.
     ::如果一个对象的加速度为零,是否意味着它的速度为零?举一个例子。
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  2. If an object has zero velocity, does that mean it has zero acceleration?  Give an example.
     ::如果一个对象为零速度,是否意味着它为零加速度?举一个例子。
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  3. If the acceleration of a motorboat is 4.0 m/s ² , and the motorboat starts from rest, what is its velocity after 6.0 s?
     ::如果机动艇加速度为4.0米/秒,而机动艇则从休息开始,那么在6.0秒之后的速度是多少?
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  4. The friction of the water on a boat produces an acceleration of -10. m/s ² . If the boat is traveling at 30. m/s and the motor is shut off, how long does it take the boat to slow down to 5.0 m/s?
     ::如果船只在30米/秒时行驶,发动机关闭,船需要多长时间才能减慢到5.0米/秒?