SOLUTION: An object drops from a cliff that is 150m high. The distance, d, in meters that the object has dropped at t seconds is modelled by d(t)=4.9t^2. a. Find the average rate of change

Question 26277: An object drops from a cliff that is 150m high. The distance, d, in meters that the object has dropped at t seconds is modelled by d(t)=4.9t^2.
a. Find the average rate of change of distance with respect to time from 2 s to 5 s.
b. Find the rate at which the object hits the ground to the nearest tenth.
THANK YOU.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
LET ME USE SYMOL S INSTEAD OF D FOR DISTANCE
S=4.9T^2
RATE OF CHANGE OF DISTANCE (MORE APPROPRIATE WORD IS RATE OF CHANGE OF
POSITION OR RATE OF DISPLACEMENT IN PHYSICS) IS VELOCITY.HOPE YOU ARE
TAUGHT DIFFERENTIATION..
V=DS/DT=4.9*2*T=9.8T
AT T= 2 SEC.,V=9.8*2=19.6 M/SEC.
AT T=5 SEC....V=4.9*5=24.5 M/SEC.
AVERAGE VELOCITY DURING THIS PERIOD=(19.6+24.5)/2=22.05 M/SEC.
S=4.9T^2=150 FOR THE OBJECT TO HIT THE GROUND
T=5.53 SEC
AT T= 5.53 SEC. V= 4.9*5.53=27.11 M/SEC.

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