Question

1/66UtuSet backgroundClear framed750СB500250Do0At what time(s) is the car stopped?100200300400500600CDuring which segment was he runningthe fastest?Al what time(s) did the car have thegreatest velocitybWhat was the farthest distance that hereached?segment A750m200to300sWhat was the greatest velocity of thecar?C2d.Al what time(s) was the caraccelerating?dWhat was his displacement between100 and 300 seconds?d=250mCDWhat was his velocity during each ofthe labeled segments?How far did he car travel between 2seconds and 4 seconds?АBCHow far did the car travel betweenand 2 seconds ?DWhat was the total distance that heo

Answer 1

Step-by-step explanation:

The graph is a displacement vs time graph. It means that the slope of the graph is velocity

a) The speed of the car is equivalent to the slope of the line. The segment with the greatest slope has the greatest speed. Segment AB has the greatest slope. Thus,

he was running fastest in segment A

b) The farthest distance is the highest value of the graph on the y axis. Thus,

Farthest distance reached = 750 m

c) The segment where the slope is zero represents the period of rest. This is segment C. Thus, the time segment for rest is 200 to 300 seconds

d) At 100 seconds, displacement = 500m

At 300 seconds, displacement = 750 m

Displacement between 100s and 300s = 750 - 500 = 250 m

e) We would find the slope of each segment. The formula for calculating slope is expressed as

m = (y2 - y1)/(x2 - x1)

where

y1 and y2 are the y coordinates of selected initial and final points on the line.

x1 and x2 are the x coordinates of the selected initial and final points on the line.

For segment A,

x1 = 0, y1 = 0

x2 = 100, y2 = 500

Velocity = (500 - 0)/(100 - 0) = 500/100

Velocity = 5 m/s

For segment B,

x1 = 100, y1 = 500

x2 = 200, y2 = 750

Velocity = (750 - 500)/(200 - 100) = 250/100

Velocity = 2.5 m/s

For segment C,

x1 = 200, y1 = 750

x2 = 300, y2 = 750

Velocity = (750 - 750)/(300 - 200) = 0/100

Velocity = 0 m/s

For segment D,

x1 = 300, y1 = 750

x2 = 500, y2 = 0

Velocity = (0 - 750)/(500 - 300) = - 750/200

Velocity = - 3.75 m/s

Total distance is the total area under the graph

Section A is a triangle whose base is 100 and height is 500.

Area of triangle = 1/2 x base x height

Area of section A = 1/2 x 100 x 500 = 25000 m

Section B is a trapezoid with the given information;

height, h = 100

opposite sides, a and b are 500 and 750

Area of trapezoid = 1/2(a + b)h

Area of section B = 1/2(500 + 700)100 = 60000

Section C is a rectangle. Its width is 100 and the height is 750

Area = 100 x 750 = 75000

Section D is a triangle. Height = 750 and base = 500 - 300 = 200

Area = 1/2 x 200 x 750 = 75000

Total distance = 25000 + 60000 + 75000 + 75000

Total distance = 235000 m