MM_F5_6 Gradient_Area Under A Graph_2

22/04/2020

斜度 面积

Chapter 6 Gradient And Area Under A Graph

6.1 Quantity represented by the gradient of a graph

- Distance-time graph

Distance Distance

Time Time

一个物体移动

 A distance-time graph can tell us how an object moves with respect to time.

 The gradient of a distance-time graph is the rate of change of distance,
速度
斜度=速度
which is the speed.

斜度越陡峭（直），速度越快

 The steeper is the gradient, the faster is the speed.

固定/休息 斜度= 0

 It is stationary if it has zero gradient (horizontal).

倒反方向

 The negative gradient indicates that it is travelling in the opposite direction,

going back to its original position.

回去原来的位置

Diagram 1 shows a object moves from O to C for a distance of 50m in a period
of 10 seconds.

Distance (m)

50 C

35 A B

O3 8 10 Time (s)

OA – positive gradient

– motion at uniform speed移动的速度一致

– motion for a distance 35 metres in a period of 3 seconds

AB – zero gradient距离没有变动 停止/固定/休息

– no change in distance means motion stops (stationary/rest)
– stationary for the period of 5 seconds (8 – 3 = 5)

BC – positive gradient

– motion continues to C 继续移动去到 C

– motion for a distance 15 metres (50 – 35 = 15) in a period of 2
seconds (10 – 8 = 2)

全程用 10 秒移动 50 米

OC – motion of 50 metres in a period of 10 seconds

Diagram 2 shows a object moves from O to R passing through P and Q.

Distance (m)

64 P Q

O4 8 R Time (s)
14

OP – positive gradient

– motion at uniform speed 移动的速度一致

– motion for a distance 64 metres in a period of 4 seconds

PQ – zero gradient 距离没有变动 停止/固定/休息

– no change in distance means motion stops (stationary/rest)
– stationary for the period of 4 seconds (8 – 4 = 4)

QR – negative gradient 移动回去原来的位置/倒转方向移动

– negative speed shows object moves back to original place or moves in
the opposite direction.

– motion for a distance 64 metres in a period of 6 seconds (14 – 8 = 6)

全程用 14 秒移动 128 米

OR – motion of 128 metres (64 x 2) in a period of 14 seconds

Example:
The distance-time graph shows the motion of a car for a period of 4.5 hours.

Distance (km) (a) Determine
270 (i) the duration when the car is
150 stationary.
(ii) the speed, in kmh–1, of the car in
0 1 2 3 45 the first two hour.

(b) Calculate the speed of the car for the

last 1.5 hours.

Time Speed = Distance
(hours) Time

(a) (i) Stationary period = 3 – 2 = 1 hour Rate of change

(ii) Speed of car in the first hour  150  0  y2  y1
20 x2  x1

 75 kmh1 这两个 formula 是
相同意思的哦！
270 150
(b) Speed of car in the last 1.5 hours  4.5  3

 80 kmh1