does the function D=16t^2 accurately represent the distance a stone fell after a given time t?
+36916 IF the initial velocity is ZERO. And you assume that the object starts at D=0 And D is measured in FEET
AND time is in SECONDS.
If it's due to the force of Earth's gravity,no. That would be D= 5(t^2). because s = 1/2a(t^2) ,that is displacement (or distance from origin) = acceleration x time if the object is starting from rest.
Acceleration due to Earth's gravity = approx 10 ms^(-2) and the D= 5(t^2) comes from the formula for displacement which is
s = ut + 1/2 a(t^2) . If u,which is the initial velocity,is zero,you get this expression,which is for an object in free fall in earth's gravity,starting from rest and ignoring air resistance.
For more on this,google SUVAT equations.
No, the earths gravitational accelerator is around 9.8. Because D=(A*T2)/2 then 9.8/2=4.9. So the accelerator is 4.9*T2
+36916 Alright you 'metric' guests....the accelaration of gravity is ALSO 32 ft/sec^2 (hence my nswer detailing that D would be in feet!)
