Physics
posted by BT .
A turtle is moving with a constant acceleration along a straight ditch. He starts his stopwatch as he passes a fence post and notes that it takes him 10s to reach a pine tree 10 m farther along the ditch., As he passes the pine tree, his speed is 1.2m/s. How far was he from the fence post when he started from rest?

Let
D1 = distance of the fence post from where the turtle started
D2 = distance of the fence post to the pine tree = 10 m (given)
Working formula is
(V2)^2  (V1)^2 = 2a(D2)
where
V2 = velocity upon passing the pine tree = 1.2 m/sec.
V1 = velocity upon passing the fence post
a = acceleration
D2 = 10 m
Substituting values,
1.2^2  (V1)^2 = 2a(10)
1.44  (V1)^2 = 20a  call this Equation A
Next working formula is
D2 = (V1)T + (1/2)aT^2
10 = (V1)(10) + (1/2)(a)(10)^2
10 = (V1)(10) + 50a
Simplifying the above,
1 = V1 + 5a
Solving for "a"
5a = 1  V1
a = (1  V1)/5  call this Equation B
Substituting Equation B into Equation A,
1.44  (V1)^2 = 20(1  V1)/5
1.44  (V1)^2 = 4  4(V1)
Modifying the above,
(V1)^2  4(V1) + 2.56 = 0
Using the quadratic formula,
V1 = 0.8 m/sec. and V1 = 3.2 m/sec.
Since the turtle is constantly accelerating, then its speed V1 at the fence post must be lower than its speed when passing the pine tree. Thus being said, the root V1 = 3.2 will be ignored.
Hence, V1 = 0.8 m/sec.
and using Equation B (to solve for "a"),
a = (1  0.8)/5
a = 0.04 m/sec^2
To solve for D1, the formula is
(V1)^2  (Vo)^2 = 2a(D1)
Since the turtle started from rest, Vo = 0, hence
0.8^2  0 = 2(0.04)(D1)
Solving for "D1"
D1 = 0.64/0.08
D1 = 8 meters