题目

Problem

The speed-time graph in Figure 2 illustrates the motion of a car travelling along a straight horizontal road.

At time $t=0$ , the car starts from rest and accelerates uniformly for $30\,\mathrm{s}$ until it reaches a speed of $V\,\mathrm{m\,s^{-1}}$ .

The car then travels at a constant speed of $V\,\mathrm{m\,s^{-1}}$ until time $t=T$ seconds.

(a) Show that the distance travelled by the car between $t=0$ and $t=T$ seconds is $V(T-15)$ metres.

(2)

A motorbike also travels along the same road.

The motorbike starts from rest at time $t=10\,\mathrm{s}$ and accelerates uniformly for $40\,\mathrm{s}$ .
The acceleration of the motorbike is the same as the acceleration of the car.
The motorbike then travels at a constant speed for a further $10\,\mathrm{s}$ before decelerating uniformly until it reaches a speed of $V\,\mathrm{m\,s^{-1}}$ at time $T$ seconds.

(b) On Figure 2, sketch a speed-time graph for the motion of the motorbike.

[If you need to redraw your sketch, there is a copy of Figure 2 on page 15.]

(2)

At time $t=T$ seconds, the distance travelled by each vehicle is the same.

(d) Find the value of $T$ .

(5)

Only use this copy of Figure 2 if you need to redraw your sketch.

题目中文翻译

图 2 中的速率—时间图像表示一辆汽车沿水平直路行驶的运动过程。

在 $t=0$ 时，汽车从静止出发，匀加速运动 $30\,\mathrm{s}$ ，直至速率达到 $V\,\mathrm{m\,s^{-1}}$ 。

随后汽车以恒定速率 $V\,\mathrm{m\,s^{-1}}$ 行驶至 $t=T$ 秒。

(a) 证明汽车在 $t=0$ 至 $t=T$ 秒之间行驶的路程为 $V(T-15)$ 米。

(2)

一辆摩托车也沿同一条公路行驶。

摩托车在** $t=10\,\mathrm{s}$ 时**从静止出发，并匀加速运动 $40\,\mathrm{s}$ 。
摩托车的加速度与汽车的加速度相同。
随后摩托车以恒定速率再行驶 $10\,\mathrm{s}$ ，然后匀减速运动，直至在 $T$ 秒时速率达到 $V\,\mathrm{m\,s^{-1}}$ 。

(b) 在图 2 上画出摩托车运动的速率—时间图像草图。

[如果需要重画草图，第 15 页提供了图 2 的副本。]

(2)

在 $t=T$ 秒时，两辆车各自行驶的路程相同。

(d) 求 $T$ 的值。

(5)

仅在需要重画草图时使用此图 2 副本。

解答

(a)

解法一

思路

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总路程等于速率图下的面积，分成最初三角形与后面的长方形。

答题过程

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The distance is the area under the car’s speed-time graph:

\begin{align*} s_{\rm car}=&\,\frac12(30)V+V(T-30),\\ =&\,15V+VT-30V,\\ =&\,\boxed{V(T-15)\ \mathrm m}, \end{align*}

as required.

(b)

解法一

思路

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必须在原汽车图的同一坐标轴上加画摩托车图。两车加速度相同，所以两条上升线平行；摩托车从 $t=10$ 开始，到 $50$ 结束加速， $60$ 开始减速。

答题过程

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On the given car graph, add the motorbike graph as follows:

It starts at $(10,0)$ and rises along a straight line parallel to the car’s rising line.
Its acceleration ends at $t=50$ . Draw a horizontal segment from $t=50$ to $t=60$ , above speed $V$ .
From $t=60$ , draw a straight falling line ending at $(T,V)$ , where it meets the car’s horizontal graph.

Label times 10, 50 and 60 clearly. Keep both vehicles on the same axes; do not draw a vertical drop at $t=T$ .

(c)

解法一

思路

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汽车加速度是 $V/30$ ，摩托车以同样加速度从静止加速 $40$ 秒，因此直接用速度变化等于加速度乘时间。

答题过程

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The car’s acceleration is $V/30\ \mathrm{m\,s^{-2}}$ . The motorbike has the same acceleration for 40 s, starting from rest.

Its constant speed is therefore

v=0+\frac V{30}(40)=\boxed{\frac{4V}{3}\ \mathrm{m\,s^{-1}}},

as required.

(d)

解法一

思路

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先把摩托车的总面积写完整：加速三角形、匀速长方形、减速梯形。令两车总路程相等，再约去正数 $V$ 求 $T$ 。

答题过程

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The motorbike’s distance is

\begin{align*} s_{\rm bike}=&\,\frac12(40)\frac{4V}{3} +\frac{4V}{3}(10) +\frac12\left(\frac{4V}{3}+V\right)(T-60),\\ =&\,40V+\frac{7V}{6}(T-60),\\ =&\,\frac{7VT}{6}-30V. \end{align*}

The distances are equal at $t=T$ , so

\begin{align*} \frac{7VT}{6}-30V=&\,V(T-15),\\ \frac T6=&\,15,\\ T=&\,\boxed{90}. \end{align*}

Here $V>0$ was cancelled. The result satisfies $T>60$ , consistent with the motorbike’s deceleration interval.

解法二

思路

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也可比较两条速率图之间的面积。摩托车在前 $40$ 秒落后的路程，等于它之后多行驶的路程，这样不用计算两车全部面积。

答题过程

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The motorbike reaches speed $V$ at

t=10+\frac{V}{V/30}=40\ \mathrm s.

Up to $t=40$ , the car travels $15V+10V=25V$ metres, while the motorbike travels $\tfrac12(30)V=15V$ metres. The motorbike’s distance deficit is therefore $10V$ metres.

From $t=40$ onwards, the area between the motorbike’s graph and the car’s horizontal graph is a trapezium. Its height is $4V/3-V=V/3$ , and its parallel time lengths are $T-40$ and $60-50=10$ . Thus the extra distance is

\frac12\left(\frac V3\right)\bigl[(T-40)+10\bigr]=\frac V6(T-30).

Equality of the total distances requires

\frac V6(T-30)=10V,\qquad T-30=60.

Hence $\boxed{T=90}$ .

M1 2023 June Q5

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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解法二

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