Answer:
Part a) The acceleration in the first 20 seconds is [tex]0.35\ m/sec^2[/tex]
Part b) The distance between the two stops is 945 meters
Part c) The average speed during the journey is 5.9 m/sec
Step-by-step explanation:
Part a)
Let
x ---> the time in seconds
y ---> the velocity in m/sec
a ---> acceleration in m/ sec^2
we know that
The acceleration is equal to divide the velocity by the time
so
The acceleration is the slope of the linear equation of the graph
For the first 20 seconds
take the points
(0,0) and (20,7)
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
substitute
[tex]m=\frac{7-0}{20-0}[/tex]
[tex]m=\frac{7}{20}=0.35\ m/sec^2[/tex]
therefore
The acceleration in the first 20 seconds is [tex]0.35\ m/sec^2[/tex]
Part b) we know that
Looking at the graph
The two stops of the bus are the points (0,0) and (160,0)
The stops of the bus represent the x-intercepts of the linear equation
Values of x when the value of y is equal to zero
In the context of the problem
Values of the time when the velocity of the bus is equal to zero
To find out the distance, we need to determine the area of the graph
(because the distance is equal to multiply the velocity by the time)
The area of the graph in the interval [0,160] is equal to the area of two right triangles plus the area of rectangle
so
[tex]A=\frac{1}{2}(20)( 7)+(130-20)(7)+ \frac{1}{2}(160-130)(7)[/tex]
[tex]A=70+770+105=945\ m[/tex]
Part c) we know that
The average speed is equal to divide the total distance by the total time
so
[tex]\frac{945}{160}=5.9\ m/sec[/tex]
