An object dropped from rest from the top of a tall building on Planet X falls a distance d (t )equals 17 t squared feet in the first t secon

Question

An object dropped from rest from the top of a tall building on Planet X falls a distance d (t )equals 17 t squared feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from t1= 6 to t2=10. This rate is known as the average​ velocity, or speed.

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Khải Quang 6 months 2021-07-15T02:40:34+00:00 1 Answers 41 views 0

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    2021-07-15T02:42:31+00:00

    Answer:

    The average speed of the object is 272 m/s.    

    Explanation:

    Given that,

    An object dropped from rest from the top of a tall building on Planet X falls a distance of :

    d(t)=17t^2

    t is time in seconds

    We need to find the average rate of change of distance with respect to time as t changes from t = 6 s to t =10 s. The rate of change of distance wrt time is called velocity of an object. Average velocity is given by :

    v(t)=\dfrac{d_{10}-d_6}{10-6}\\\\v(t)=\dfrac{17(10)^2-17(6)^2}{10-6}\\\\v(t)=272\ m/s

    So, the average speed of the object is 272 m/s.

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