Calculate Acceleration of Falling Hammer on Moon | 1-D Kinematics Problem

[RubenL]

Homework Statement

An astronaut standing on a platform on the Moon
drops a hammer. If the hammer falls 6.0 meters vertically in 2.7 seconds, what is its acceleration?

Homework Equations

x=x0+v0t+(1/2)at²

The Attempt at a Solution

x=x0+v0t+(1/2)at²
6=0+.5(a)(2.7²)
6=3.7a
a=1.62m/s²

This is a very simple question and easily solvable...however my question is...considering this kinematics equation x=x0+v0t+(1/2)at².
I can understand how all of the variables play within this equation, but i do not understand where the 1/2 comes from (i understand it has to be this way, however), what is the purpose of halving you answer?

(this is wrong, but) why can't the equation be solved like so;

X=6m
Vi=0m/s
t=2.7s
a=?m/s²

6m / 2.7s = 2.2m/s
2.2m/s / 2.7s = .81m/s²

and leave it like so? (like i said i know this is wrong, but i am just trying to figure out why the 1/2 is in the equation)

 

[climb515c]

Im not very good at typing in the formulas. But the 1/2 comes from the derivation of the expression you are using, and the properties of calculus.

x=x_o+V_ot +1/2at²

is derived from v=dx/dt, separate the dx and dt (v*dt=dx) and integrate both sides. You can subsitute v for (v_o +at), since you velocity at any time is equal to your initial velocity + your accel*time.

so you now are integrating both sides of dx = (v_o +at)dt. When you integrate the second term on the right "a*t", a is constant, so you can pull it out of the integral, then integrate t*dt, which is 1/2*t². the first term on the right side integrates to v_o*t

this whole integral gives you x=v_ot +1/2at² +C

When t=0, you find C=x_o

so the final expression you are using is x=x_o+V_ot +1/2at²

hopefully that helps

 

[RubenL]

Im not very good at typing in the formulas. But the 1/2 comes from the derivation of the expression you are using, and the properties of calculus.

x=x_o+V_ot +1/2at²

is derived from v=dx/dt, separate the dx and dt (v*dt=dx) and integrate both sides. You can subsitute v for (v_o +at), since you velocity at any time is equal to your initial velocity + your accel*time.

so you now are integrating both sides of dx = (v_o +at)dt. When you integrate the second term on the right "a*t", a is constant, so you can pull it out of the integral, then integrate t*dt, which is 1/2*t². the first term on the right side integrates to v_o*t

this whole integral gives you x=v_ot +1/2at² +C

When t=0, you find C=x_o

so the final expression you are using is x=x_o+V_ot +1/2at²

hopefully that helps

Great! This helps put reason to it. I see i need to get my head into calculus to fully understand. But its clear to me now that there is a mathematical reason behind it.