Question

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.

Answer 1

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