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Salsk061 [2.6K]
3 years ago
13

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

a while, and finally runs quickly back to her initial position along a straight line. Which of the following statements is true about the average speed and the magnitude of the average velocity of the student during her trip? Her average speed is greater than the magnitude of her average velocity. Her average speed is the same as the magnitude of her average velocity. Her average speed is less than the magnitude of her average velocity. The student's average speed and velocity cannot be compared without knowing the farthest point she reached.
Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

Answer:Average speed is greater than average velocity.

Explanation  :

Given

student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position

Let x be the length of track and the whole process takes t time

For average speed =\frac{distance\ traveled}{time\ taken}

Average speed=\frac{2x}{t}

For average velocity =\frac{Displacement}{time\ taken}

Since displacement is zero as she returns to its initial position.

Average velocity=0

Therefore Average speed is greater than average velocity.

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Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
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the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

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