does the function D=16t^2 accurately represent the distance a stone fell after a given time t?

Guest Feb 17, 2017

#1**0 **

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.

ElectricPavlov
Feb 17, 2017

#2**0 **

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.

Guest Feb 17, 2017

#3**0 **

No, the earths gravitational accelerator is around 9.8. Because D=(A*T^{2})/2 then 9.8/2=4.9. So the accelerator is 4.9*T^{2}

Guest Feb 17, 2017

#4**0 **

Alright you 'metric' guests....the accelaration of gravity is ALSO 32 ft/sec^2 (hence my nswer detailing that D would be in feet!)

ElectricPavlov
Feb 18, 2017