Skip to content
CalcGospel 國際數學圖譜
返回

M1 2019 Oct Q6

A Level / Edexcel / M1

IAL 2019 Oct Paper · Question 6

题目

Problem

6.

An athlete runs a 200 m200\,\mathrm{m} race along a straight horizontal track.

In a model of the motion of the athlete, air resistance is ignored, the athlete starts from rest at time t=0t=0 seconds and moves with uniform acceleration 0.8 m s−20.8\,\mathrm{m\,s^{-2}} for TT seconds, reaching a speed of V m s−1V\,\mathrm{m\,s^{-1}}. She then maintains this speed until she crosses the finishing line.

The total time from when the athlete starts to when she crosses the finishing line is 30 s30\,\mathrm{s}.

(a) Sketch a speed-time graph for the model of the motion of the athlete from the instant when she starts to the instant when she crosses the finishing line.

(2)

(b) Write down an expression for VV in terms of TT.

(1)

(c) Show that T2−kT+500=0T^2-kT+500=0, where kk is a constant to be found.

(4)

(d) Hence find the value of TT, justifying your answer carefully.

(3)

(e) Considering your speed-time graph or otherwise, state two ways, apart from including air resistance, in which the model could be made to be more realistic.

(2)
题目中文翻译

一名运动员沿水平直跑道参加 200 m200\,\mathrm{m} 赛跑。

在她的运动模型中,忽略空气阻力。运动员在 t=0t=0 秒时从静止出发,以恒定加速度 0.8 m s−20.8\,\mathrm{m\,s^{-2}} 运动 TT 秒,速率达到 V m s−1V\,\mathrm{m\,s^{-1}}。随后,她保持这一速率,直至越过终点线。

从起跑到越过终点线的总时间为 30 s30\,\mathrm{s}。

(a) 为该模型画出运动员从起跑瞬间到越过终点线瞬间的速率—时间图像草图。

(2)

(b) 写出用 TT 表示 VV 的表达式。

(1)

(c) 证明 T2−kT+500=0T^2-kT+500=0,其中 kk 是一个待求的常数。

(4)

(d) 由此求 TT 的值,并仔细说明答案的理由。

(3)

(e) 结合你的速率—时间图像或用其他方法,指出除考虑空气阻力外,使模型更符合实际的两种改进方式。

(2)

解答

(a)

解法一

思路

展开

先匀加速再匀速,图像是一条从原点出发的上升直线,接一段水平线。到终点时仍有速率 VV,不能画出停车的竖直实线。

答题过程

展开

Label the horizontal axis tt (seconds) and the vertical axis vv (m s−1\mathrm{m\,s^{-1}}). Join

(0,0) to (T,V) to (30,V)(0,0)\ \text{to}\ (T,V)\ \text{to}\ (30,V)

by straight segments. The first segment rises, and the second is horizontal. Label TT, 3030 and VV. The graph ends at (30,V)(30,V) without a solid vertical drop to the time axis.

(b)

解法一

思路

展开

第一段由静止开始,用末速度等于加速度乘时间。

答题过程

展开

Using v=u+atv=u+at for the acceleration stage,

V=0.8T.\boxed{V=0.8T}.

(c)

解法一

思路

展开

整段路程等于图像下三角形与矩形的面积和。承接 (b) 代入 VV,展开后整理成题目所给的二次方程。

答题过程

展开

The area under the speed-time graph is 200 m200\,\mathrm{m}, so

12TV+(30−T)V=200.\frac12TV+(30-T)V=200.

Substituting V=0.8TV=0.8T,

0.4T2+24T−0.8T2= 200,−0.4T2+24T= 200,T2−60T+500= 0.\begin{align*} 0.4T^2+24T-0.8T^2=&\,200,\\[4mm] -0.4T^2+24T=&\,200,\\[4mm] T^2-60T+500=&\,0. \end{align*}

This is the required form, with k=60\boxed{k=60}.

(d)

解法一

思路

展开

解二次方程得到两个代数根后,要用加速段包含在全程 3030 秒以内的条件排除不合理的根。

答题过程

展开

From (c),

T2−60T+500=(T−10)(T−50)=0.T^2-60T+500=(T-10)(T-50)=0.

Hence T=10T=10 or T=50T=50. Since the acceleration stage is followed by a constant-speed stage within a total time of 30 s30\,\mathrm{s}, we require 0<T<300<T<30.

Therefore T=50T=50 is inadmissible and T=10 s\boxed{T=10\,\mathrm{s}}.

(e)

解法一

思路

展开

只给两项明确的模型改进,分别针对起跑段恒定加速度和后段恒定速率。不要重复题目已排除的空气阻力。

答题过程

展开
  1. Allow the acceleration to vary during the initial stage, rather than being constant.
  2. Allow the speed to fluctuate during the later stage, rather than remaining exactly constant.