Respuesta :
Answer:
Step-by-step explanation:
A. for BC, 40-20=20
4-0=4
20/4=5km/h
for CD, speed=0 as it stops
for DE, it is // to BC so speed also 5km/h
B. stop @ C when t=4
arrive @ E when t=10
10-4 = 6h
C. BC is d=40-5t
CD is d=20
DE is d=20-5(t-6)=50-5t
The speed at a point of a distance to time graph is the same as the slope of the graph at that point
The correct values of the motion are;
A) Speed of the pedestrian at BC is -5 km/h
Speed of the pedestrian at CD is 0 km/h
Speed of the pedestrian at DE is -5 km/h
B) The time after the stop it took the pedestrian to arrive point E is 4 hours
C) At BC, d(t) = -5·t + 40
At CD d(t) = 20
At DE d(t) = 20 - 5·x
The reason the above values are correct are given as follows;
The given graph is a distance (d, km) to time (t, h)
A) The speed of an object with motion given on a distance to time graph is the slope of the graph at a specified point
The coordinates of point on the graph are;
B(0, 40), C(4, 20), D(6, 20), and E(10, 0)
The speed between points B and C is therefore;
[tex]Speed \ BC = \dfrac{20 \, km - 40 \, km}{4 \, h - 0 \, h } = -5 \, km/h[/tex]
The speed of the pedestrian along BC = -5km/h
The speed between points C and D is given as follows;
[tex]Speed \ CD = \dfrac{20 \, km - 20 \, km}{6 \, h - 4 \, h } = 0 \, km/h[/tex]
The speed of the pedestrian along CD is 0 km/h
The equation to find the speed between points D and E is therefore;
[tex]Speed \ DE = \dfrac{20 \, km - 0 \, km}{10 \, h - 6 \, h } = -5 \, km/h[/tex]
The speed of the pedestrian along DE = -5km/h
B) A stop is a point of zero speed
The stop is between the interval CD
Time at which the pedestrian start moving again is point D, which is 6 hours after start
Time at which the pedestrian arrived at point E is 10 hours after start
Therefore, the time since the stop at which the pedestrian arrive at point E is 10 hours - 6 hours = 4 hours
C) The formula for the function d(t) are given in the form y = m·x + c are found as follows;
The equation of the distance, d(t) = m·x + c
Where;
m = The slope of the line = The speed
c = The initial value of the distance (initial, d-value)
Between point B and C, we have;
The slope, m = The speed = -5 km/h
The initial value of the distance , c = 40
∴ The formula for d(t) for section BC is d(t) = -5·t + 40 (0 ≤ t ≤ 4)
Between point C and D
The speed, m = 0
The initial y-value = The 40 + (-5) × 4 = 20
The function for d(t) for the section CD is d(t) = 20 + 0 × x = 20
Therefore, the function is, d(t) = 20 (4 ≤ t ≤ 6)
Between point D and E
The speed, m = -5 km/h
The initial y-value = 20 km
The function for d(t) for the section for the section DE is d(t) = 20 - 5·x (6 ≤ t ≤ 10)
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