Question 26277
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.