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The more detailed solving process, the better. And hope you can make clear handwriting. Thank you

1. Explain why the average velocity found in Example 1 is larger than the instantaneous velocity found in Example 2. (Hint: Sketch the graph of velocity as a function of time. Is the function increasing or decreasing? What does that tell you about how the velocities at the beginning and end of the time interval compare with the average velocity during that time interval?)

Example 1: Consider a car which has a position given by x)-15.0m +2.0^/. To calculate the average velocity between 3.0 s and 7.0 s, we first calculate the position at 3.0 s and at 7.0s. x(3.0s) 15.0m +2.0 (3.0s) 33m (7.0s)-5.0m+2.03(7.0s) -113m The average velocity of the car between t 3.0 s and t - 7.0s is 7.0)3.0)113m-33m-20- 7.0s-3.0s 10s

Example 2: Consider the car in Example 1. What is the instantaneous velocity of the car at t 3.O s? We answer this by first finding the velocity as a function of time. dx d dt dit Note that the units on the constants do not change when we take a derivative and they will provide a check on our work. Now to get the instantaneous velocity at t - 3.0 s, we evaluate our velocity function for t 3.0 s. v(3.0 s)-4.0% , 3.0 s-12

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Answer #1

Function is increasing , because velocity increases with time.

5t8 : 33 m 113 m Auerage velouty bistana covered by Ca Di人卜an.u covered Time taken ,,3-33 熙 4 ⓢ s.utanyan n4 vetwy 쓿 dlt dt at time 3 sec at time .ユ see. じ. + V aj 12+ 28 2.

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