High School

The graph shows the distance of a car traveling along a straight road for 8 hours. A positive velocity indicates motion to the right, and a negative velocity indicates motion to the left.

11. Determine the rates of change in distance with respect to time.

12. Between what two times is the car not moving?

13. Between what two times is the car traveling to the right?

14. Between what two times is the car traveling to the left?

15. Between what two times is the car traveling the fastest?

16. What does the x-intercept (8, 0) represent in this situation?

Answer :

The graph of the distance traveled by the car to the time taken during the 8 hours along the straight road gives;

11. The rates of change between the following times, t, are;

  • t = 0 and t = 2 the rate of change is 30 mph
  • t = 2 and t = 3 the rate of change is 60 mph
  • t = 3 and t = 4 the rate of change is 0 mph
  • t = 4 and t = 8 the rate of change is -30 mph

12. The car is not moving between times 3 and 4 hours

13. The car is traveling to the right between times 0 and 3 hours

14. The car is traveling to the left between times 4 and 8 hours

15. The car is traveling fastest between times 2 and 3 hours

What is a distance–time graph?

A distance–time graph is one that shows the variation of the distance traveled with respect to time, such that a positive distance traveled indicates a forward distance and a negative distance traveled indicates reversed motion.

The points on the graph in the possible question obtained online are;

(0, 0), (2, 60), (3, 120), (4, 120), (8, 0)

From the above points, the attached graph is constructed using Sheets

11. The rates of change in distance with respect to time are;

  • [tex] \displaystyle { \frac{ (\triangle d)_{1} }{ \triangle t} = \frac{60 - 0}{2 - 0} = 30}[/tex]; Velocity = 30 mph
  • [tex] \displaystyle { \frac{ (\triangle d)_{2} }{\triangle t} = \frac{120 - 60}{3 - 2} = 60}[/tex]; Velocity = 60 mph
  • [tex] \displaystyle { \frac{ (\triangle d)_{3} }{\triangle t} = \frac{120 - 1200}{4 - 3} = 0}[/tex]; Velocity = 0 mph
  • [tex] \displaystyle { \frac{ (\triangle d)_{4} }{\triangle t} = \frac{0 - 120}{8 - 4} = - 30}[/tex]; Velocity= -30 mph

12) When the car is not moving, we have that the rate of change of distance with time is zero, that is the car's distance does not change with time.

The two times between which the car is not moving is between the 3 hour and 4 hour points on the graph

13. The velocity of the car is given by the rate of change of the distance with time of the car

The positive rate of change on the graph is given between the times 0 and 3 hours after the car starts motion.

The car is traveling to the right between times, t = 0 and t = 3

14. The car has negative rate of change and therefore, negative velocity, and therefore also the car is moving to the left between times, t = 4 hours, and t = 8 hours

15. The car is traveling fastest when the rate of change of the car's distance traveled with time is highest, which is given by the times between t = 2 hours and t = 3 hours in which the car is traveling at 60 miles per hour

Learn more about distance time graph here;

https://brainly.com/question/13877898

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