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skelet666 [1.2K]
3 years ago
11

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 parenth

esis 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.
Physics
1 answer:
melamori03 [73]3 years ago
4 0

displacement is given by equation

d = 18t^2

now at t = 5 s the position is

d_1 = 18 *5^2 = 450 m

similarly position at t = 9 s

d_2 = 18*9^2 = 1458 m

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

d = 1458 - 450 = 1008 m

time interval

\delta t = 9 - 5 = 4 s

now the average velocity will be given as

v = \frac{\delta x}{\delta t}

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

so its average speed is 252 m/s

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