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An object dropped from rest from the top of a tall building on planet x falls a distance d(t)18 left parenthesis t right parenthesis equals 18 t squaredd(t)=18t^2 feet in the first t seconds. find the average rate of change of distance with respect to time as t changes from t 1t1equals=55 to t 2t2equals=99. this rate is known as the average​ velocity, or speed.

Respuesta :

displacement is given by equation

[tex]d = 18t^2[/tex]

now at t = 5 s the position is

[tex]d_1 = 18 *5^2 = 450 m[/tex]

similarly position at t = 9 s

[tex]d_2 = 18*9^2 = 1458 m[/tex]

so the displacement of object in given interval of time will be

[tex]d = 1458 - 450 = 1008 m[/tex]

time interval

[tex]\delta t = 9 - 5 = 4 s[/tex]

now the average velocity will be given as

[tex]v = \frac{\delta x}{\delta t}[/tex]

[tex]v = \frac{1008}{4} = 252 m/s[/tex]

so its average speed is 252 m/s

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