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April 16, 2014

April 16, 2014

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

- Physics -
**a**, Wednesday, February 8, 2012 at 7:06pmLet

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

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