Question

Starting from rest, a boulder rolls down a hill with constant acceleration and travels 4.00 m during the first second.

A.) How far does it travel during the second second? d=?
B.)How fast is it moving at the end of the first second? v=?
C.)How fast is it moving at the end of the second second? v=?

Answer 1

vf^2 = vi^2 + 2*a*x



vf = at + v0
v0 = 0
vf = a*t
t = 1
vf = a

a^2 = 2*a*x

a^2 = 2a*4
a^2 = 8a
a^2 -8a = 0
a*(a - 8) = 0

a = 8 m/s2

x(t) = 4t^2
x(1) = 4
x(2) = 16

16 -4 = 12 meters
B)
v = at + v0
v = at
v = 8t
v = 8 m/s
C)
v = 8t
v = 16 m/s
PLEASE RATE.