VELOCITY AND RATES OF CHANGE
A test tube is knocked off a tower at the top of the Green building. (For the purposes of this experiment the tower is 400 feet above the ground, and all the air in the vicinity of the Green building was evacuated, so as to eliminate wind resistance.) The test tube drops 16t2[16 t-squared] feet in t seconds. Calculate: (a) the instantaneous speed at landing
PLEASE NOTE: DON'T CHANGE UNITS. ANSWER IN (ft/sec).
The answer is -160.
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1 solution
[h(t) − h(5)]/[t-5]
={ 400 − 16t2-0}/{t-5}
={16(5 − t)(5 + t)}/{t − 5}
=−16(5 + t) → −160 ft/sec as t → 5
PLEASE NOTE: (16t2)-> [16 t-squared].