Question

Answer with all calculations

Answer 1

Answer:

Part a) The acceleration in the first 20 seconds is $0.35\ m/sec^2$

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

$m=\frac{y2-y1}{x2-x1}$

substitute

$m=\frac{7-0}{20-0}$

$m=\frac{7}{20}=0.35\ m/sec^2$

therefore

The acceleration in the first 20 seconds is $0.35\ m/sec^2$

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

$A=\frac{1}{2}(20)( 7)+(130-20)(7)+ \frac{1}{2}(160-130)(7)$

$A=70+770+105=945\ m$

Part c) we know that

The average speed is equal to divide the total distance by the total time

so

$\frac{945}{160}=5.9\ m/sec$