CalcGospel 國際數學圖譜
返回

M1 2021 Oct Q3

A Level / Edexcel / M1

IAL 2021 Oct Paper · Question 3

题目

Problem

3.

A car is moving at a constant speed of 25 m s−125\,\mathrm{m\,s^{-1}} along a straight horizontal road.

The car is modelled as a particle.

At time t=0t=0, the car is at the point AA and the driver sees a road sign 48 m48\,\mathrm{m} ahead.

Let tt seconds be the time that elapses after the car passes AA.

In a first model, the car is assumed to decelerate uniformly at 6 m s−26\,\mathrm{m\,s^{-2}} from AA until the car reaches the road sign.

(a) Use this first model to find the speed of the car as it reaches the sign.

(2)

The road sign indicates that the speed limit immediately after the sign is 13 m s−113\,\mathrm{m\,s^{-1}}.

In a second model, the car is assumed to decelerate uniformly at 6 m s−26\,\mathrm{m\,s^{-2}} from AA until it reaches a speed of 13 m s−113\,\mathrm{m\,s^{-1}}. The car then maintains this speed until it reaches the road sign.

(b) Use this second model to find the value of tt at which the car reaches the sign.

(4)

In a third model, the car is assumed to move with constant speed 25 m s−125\,\mathrm{m\,s^{-1}} from AA until time t=0.2t=0.2, the car then decelerates uniformly at 6 m s−26\,\mathrm{m\,s^{-2}} until it reaches a speed of 13 m s−113\,\mathrm{m\,s^{-1}}. The car then maintains this speed until it reaches the road sign.

(c) Use this third model to find the value of tt at which the car reaches the sign.

(4)
题目中文翻译

一辆汽车沿水平直路以恒定速率 25 m s−125\,\mathrm{m\,s^{-1}} 行驶。

将汽车视为质点。

在 t=0t=0 时，汽车位于 AA 点，司机看到前方 48 m48\,\mathrm{m} 处的一个路牌。

设汽车经过 AA 点后所经历的时间为 tt 秒。

在第一个模型中，假设汽车从 AA 点开始，以恒定减速度 6 m s−26\,\mathrm{m\,s^{-2}} 减速，直至到达路牌。

(a) 使用第一个模型，求汽车到达路牌时的速率。

(2)

路牌显示，经过路牌后的限速为 13 m s−113\,\mathrm{m\,s^{-1}}。

在第二个模型中，假设汽车从 AA 点开始，以恒定减速度 6 m s−26\,\mathrm{m\,s^{-2}} 减速，直至速率达到 13 m s−113\,\mathrm{m\,s^{-1}}，然后以此速率匀速行驶至路牌。

(b) 使用第二个模型，求汽车到达路牌时 tt 的值。

(4)

在第三个模型中，假设汽车从 AA 点开始以恒定速率 25 m s−125\,\mathrm{m\,s^{-1}} 行驶至 t=0.2t=0.2，随后以恒定减速度 6 m s−26\,\mathrm{m\,s^{-2}} 减速，直至速率达到 13 m s−113\,\mathrm{m\,s^{-1}}，然后以此速率匀速行驶至路牌。

(c) 使用第三个模型，求汽车到达路牌时 tt 的值。

(4)

解答

(a)

解法一

思路

展开

第一模型从 AA 开始一直匀减速，已知距离而不需要时间，直接用速度平方公式。

答题过程

展开

Taking the direction of motion as positive, u=25 m s−1u=25\ \mathrm{m\,s^{-1}}, a=−6 m s−2a=-6\ \mathrm{m\,s^{-2}} and s=48s=48 m. Thus

v2=252+2(−6)(48)=49.v^2=25^2+2(-6)(48)=49.

The speed at the sign is 7 m s−1\boxed{7\ \mathrm{m\,s^{-1}}}.

(b)

解法一

思路

展开

第二模型分为减速和匀速两段。先算减速到 1313 的时间及距离，再用剩余距离除以 1313，不能把全程都当作匀减速。

答题过程

展开

The time to decelerate from 2525 to 13 m s−113\ \mathrm{m\,s^{-1}} is

τ=13−25−6=2 s.\tau=\frac{13-25}{-6}=2\ \mathrm s.

The distance covered during deceleration is

s1=25+132(2)=38 m.s_1=\frac{25+13}{2}(2)=38\ \mathrm m.

There are 48−38=1048-38=10 m left, travelled at 13 m s−113\ \mathrm{m\,s^{-1}}. Hence the total time is

t=2+1013=3613 s≈2.77 s.t=2+\frac{10}{13}=\boxed{\frac{36}{13}\ \mathrm s}\approx2.77\ \mathrm s.

(c)

解法一

思路

展开

第三模型多了最初 0.20.2 秒的匀速段。这一段走过 55 米，减速段仍是 22 秒和 3838 米，最后的匀速距离因此缩短。

答题过程

展开

The distance travelled during the first 0.20.2 s is

s0=25(0.2)=5 m.s_0=25(0.2)=5\ \mathrm m.

The subsequent deceleration from 2525 to 13 m s−113\ \mathrm{m\,s^{-1}} still takes 22 s and covers 3838 m.

The remaining distance is 48−5−38=548-5-38=5 m, travelled at 13 m s−113\ \mathrm{m\,s^{-1}}. Therefore

t=0.2+2+513=16865 s≈2.58 s.t=0.2+2+\frac5{13}=\boxed{\frac{168}{65}\ \mathrm s}\approx2.58\ \mathrm s.