A jogger runs 6 km north, 5 km east, then another 4 km north. Her average speed 8 km/h. How long will it take her to complete her run?10.
Using the information from the previous question, what is her average velocity during this time?

Respuesta :

Answer:

The time taken to complete her run is 1.9 hr.

Explanation:

Speed is a scalar quantity and it is defined as the ratio of distance covered to the time taken to cover that distance. As distance is also a scalar quantity, so the directions given in the problem can be ignored. Thus, the distance covered by the jogger is the sum of kilometers given in problem.

Distance covered = 6+5+4 = 15 km

And the speed is given as 8 km/hr.

So the time taken will be ratio of distance to speed.

So the jogger will take 1.9 hr to complete her run.

  1. The time taken by the jogger to complete the run is 1.875 hours
  2. The average velocity of the runner is 5.96 km/h

The given parameter:

initial position of the Jogger = 6 km north

second position of the Jogger = 5 km east

final position of the Jogger, = 4 km north

average speed of the jogger = 8km/h

To find:

  • the total time taken by the runner
  • the average velocity of the runner

Calculate the time taken for each position:

[tex]time = \frac{distance }{speed} \\\\t_1 = \frac{6 \ km}{8 \ km/h} = 0.75 \ hr\\\\t_2 = \frac{5 \ km}{8 \ km/h} = 0.625 \ hr\\\\t_3 = \frac{4 \ km}{8 \ km/h} = 0.5 \ hr\\\\[/tex]

Total time = 0.75 hr + 0.625 hr + 0.5 hr = 1.875 hr

The time (hour, minutes and seconds) = 1 hour, 52 minutes, 30 seconds

The average velocity is calculated as:

[tex]average \ velocity = \frac{displacement }{time}[/tex]

The displacement is the shortest point between the initial position and the final position.                                    

                                                         Q

                                                          ↑

                                                          ↑ 4km

                                                          ↑

                             ↑→ → → →5km→ → →

                             ↑ 6km                   |

                             ↑                            |

                             p -----------------------| Y

The line PQ is the displacement = hypotenuse side

The line PY = base of the right triangle = 5 km

The line YQ = height of the triangle = 4 km + 6km = 10 km

The line PQ is calculated as:

[tex]PQ^2 = 5^2 + 10^2\\\\PQ^2 = 125\\\\PQ = \sqrt{125} \\\\PQ = 11.18 \ km[/tex]

The average velocity is calculated as:

[tex]V = \frac{11.18}{1.875} = 5.96 \ km/hr\\\\V(m/s) = \frac{5.96}{3.6} = 1.66 \ m/s[/tex]

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