Posted by BT on Sunday, October 11, 2009 at 5:53pm.
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