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IAL 2022 June Q8

A Level / Edexcel / P3

IAL 2022 June Paper · Question 8

题目

Problem

Figure 4 is a graph showing the velocity of a sprinter during a 100 m race.

The sprinter’s velocity during the race, vv m s−1^{-1}, is modelled by the equation

v=12−et−10−12e−0.75tt≥0v=12-e^{t-10}-12e^{-0.75t}\qquad t\ge 0

where tt seconds is the time after the sprinter begins to run.

According to the model,

(a) find, using calculus, the sprinter’s maximum velocity during the race.

(5)

Given that the sprinter runs 100 m in TT seconds, such that

∫0Tv dt=100\int_0^T v\,dt=100

(b) show that TT is a solution of the equation

T=112(116−16e−0.75T+eT−10−e−10)T=\frac1{12}\left(116-16e^{-0.75T}+e^{T-10}-e^{-10}\right)
(4)

The iteration formula

Tn+1=112(116−16e−0.75Tn+eTn−10−e−10)T_{n+1}=\frac1{12}\left(116-16e^{-0.75T_n}+e^{T_n-10}-e^{-10}\right)

is used to find an approximate value for TT

Using this iteration formula with T1=10T_1=10

(c) find, to 4 decimal places,

(i) the value of T2T_2

(ii) the time taken by the sprinter to run the race, according to the model.

(3)
题目中文翻译

图 4 给出了短跑选手在 100 m 比赛中的速度图像。

选手在比赛中的速度 vv（单位：m s−1^{-1}）由方程

v=12−et−10−12e−0.75tt≥0v=12-e^{t-10}-12e^{-0.75t}\qquad t\ge 0

建模，其中 tt 秒表示选手开始跑步后的时间。

根据该模型，

(a) 用微积分求这位选手在比赛中的最大速度。

已知选手在 TT 秒内跑完 100 m，满足

∫0Tv dt=100\int_0^T v\,dt=100

(b) 证明 TT 是方程

T=112(116−16e−0.75T+eT−10−e−10)T=\frac1{12}\left(116-16e^{-0.75T}+e^{T-10}-e^{-10}\right)

的一个解。

使用迭代公式

Tn+1=112(116−16e−0.75Tn+eTn−10−e−10)T_{n+1}=\frac1{12}\left(116-16e^{-0.75T_n}+e^{T_n-10}-e^{-10}\right)

来求 TT 的近似值。

当 T1=10T_1=10 时，

(c) 求下列各值（精确到小数点后 4 位）：

(i) T2T_2

(ii) 根据该模型选手跑完全程所需的时间。

解答

(a)

We have

v=12−et−10−12e−0.75tv=12-e^{t-10}-12e^{-0.75t}

求导得

dvdt=−et−10+9e−0.75t\frac{dv}{dt} =-e^{t-10}+9e^{-0.75t}

At maximum velocity,

dvdt=0\frac{dv}{dt}=0

So

−et−10+9e−0.75t=0-e^{t-10}+9e^{-0.75t}=0

Hence

9e−0.75t=et−109e^{-0.75t}=e^{t-10}

Divide by e−0.75te^{-0.75t}:

9=e1.75t−109=e^{1.75t-10}

Taking natural logarithms,

ln⁡9=1.75t−10\ln9=1.75t-10

Therefore

t=10+ln⁡91.75t=\frac{10+\ln9}{1.75}

So

t=6.969…t=6.969\ldots

Substitute this into the expression for vv:

v=12−e6.969…−10−12e−0.75(6.969…)v=12-e^{6.969\ldots-10}-12e^{-0.75(6.969\ldots)}

Thus

v=11.887…v=11.887\ldots

The maximum velocity is

11.9 m s−1\boxed{11.9\text{ m s}^{-1}}

to 3 significant figures.

(b)

The distance travelled is

∫0Tv dt\int_0^T v\,\mathrm{d}t

Using

v=12−et−10−12e−0.75tv=12-e^{t-10}-12e^{-0.75t}

we get

∫v dt=12t−et−10+16e−0.75t\int v\,\mathrm{d}t =12t-e^{t-10}+16e^{-0.75t}

Therefore

∫0Tv dt=[12t−et−10+16e−0.75t]0T\int_0^T v\,\mathrm{d}t =\left[12t-e^{t-10}+16e^{-0.75t}\right]_0^T

Since the sprinter runs 100100 m,

[12t−et−10+16e−0.75t]0T=100\left[12t-e^{t-10}+16e^{-0.75t}\right]_0^T=100

So

12T−eT−10+16e−0.75T−(−e−10+16)=10012T-e^{T-10}+16e^{-0.75T}-(-e^{-10}+16)=100

Hence

12T−eT−10+16e−0.75T+e−10−16=10012T-e^{T-10}+16e^{-0.75T}+e^{-10}-16=100

Rearrange:

12T=116−16e−0.75T+eT−10−e−1012T=116-16e^{-0.75T}+e^{T-10}-e^{-10}

Therefore

T=112(116−16e−0.75T+eT−10−e−10)\boxed{ T=\frac1{12}\left(116-16e^{-0.75T}+e^{T-10}-e^{-10}\right) }

as required.

(c)(i)

Using T1=10T_1=10,

T2=112(116−16e−0.75(10)+e10−10−e−10)T_2=\frac1{12}\left(116-16e^{-0.75(10)}+e^{10-10}-e^{-10}\right)

So

T2=9.7493\boxed{T_2=9.7493}

to 4 decimal places.

(c)(ii)

Continuing the iteration,

T1=10,T2=9.7493,T3=9.7306,T4=9.7294,T5=9.7293\begin{aligned} T_1&=10,\\ T_2&=9.7493,\\ T_3&=9.7306,\\ T_4&=9.7294,\\ T_5&=9.7293 \end{aligned}

Therefore the time taken by the sprinter to run the race is

9.7293 s\boxed{9.7293\text{ s}}