The graph shows the movement of a car from point A to point D, there are two slant line AB and CD. (1) Calculate the speed of the car during the first 10 seconds.
(2) For how long did the car stop.
(3) What is the average speed of the car for the whole journey.
(4) Why the line BC is flat not slant.

The answer is below

Step-by-step explanation:

a) The speed of the car during the first 10 seconds is equal to the slope of the line AB. The line AB passes through (0, 0) and (10, 200)

slope = (200 - 0)/(10 - 0) = 20 m/s

The speed of the car during the first 10 seconds is 20 m/s

b) The time for the car to move from point A to point D = 60 seconds

c) Distance from point A to B = 200 m, distance from C to D = 500 - 200 = 300 m, distance from B to C = 20 m/s * 20 s = 400 m

Total distance from A to D = 200 + 400 + 300 = 900 m

Average speed = total distance / total time = 900 m / 60 seconds = 15 m/s

d) line BC is flat because the car moves at a constant speed from point B to point C. That is the speed does not change.

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: