Finding a formula...

**Danielisew** · June 10th 2007, 08:33 AM

13. The distance, D, travelled by a particle is directly proportional to the square of the time, t, taken.

When t = 40, D = 30

(a) Find a formula for D in terms of t.
(b) Calculate the value of D when t = 64
(c) Calculate the value of t when D = 12
Give your answer correct to 3 sig fig.

**Jhevon** · June 10th 2007, 08:35 AM

Originally Posted by Danielisew

13. The distance, D, travelled by a particle is directly proportional to the square of the time, t, taken.

When t = 40, D = 30

(a) Find a formula for D in terms of t.
(b) Calculate the value of D when t = 64
(c) Calculate the value of t when D = 12
Give your answer correct to 3 sig fig.

Hint

**topsquark** · June 10th 2007, 08:35 AM

Originally Posted by Danielisew

13. The distance, D, travelled by a particle is directly proportional to the square of the time, t, taken.

When t = 40, D = 30

(a) Find a formula for D in terms of t.
(b) Calculate the value of D when t = 64
(c) Calculate the value of t when D = 12
Give your answer correct to 3 sig fig.

From the problem statement $D = kt^2$ where k is some constant.

When t = 40, D = 30. So:
$D = kt^2$

$30 = k(40)^2$

$k = \frac{30}{1600} = \frac{3}{160}$

You should be able to do the rest.

-Dan

**Danielisew** · June 10th 2007, 08:41 AM

Originally Posted by topsquark

From the problem statement $D = kt^2$ where k is some constant.

When t = 40, D = 30. So:
$D = kt^2$

$30 = k(40)^2$

$k = \frac{30}{1600} = \frac{3}{160}$

You should be able to do the rest.

-Dan

What would you put for part (a) the formula?

**Jhevon** · June 10th 2007, 08:43 AM

Originally Posted by Danielisew

What would you put for part (a) the formula?

$D = \frac {3}{160}t^2$

Just replace k in the general formula Dan showed you

**Danielisew** · June 10th 2007, 08:52 AM

WAT THE.. I am confused :S

**Jhevon** · June 10th 2007, 08:57 AM

Originally Posted by Danielisew

WAT THE.. I am confused :S

Did you look at the hint I gave you?

By definition, two quantities are directly proportional if one is a constant times the other. so saying that D is proportional to t^2 is saying that

$D = k t^2$

That is the general formula for proportional quantities. Now to answer specific questions, we need to know exactly what the constant is. Dan used the two given values of D and t to find k, so now we just replace that k in the general formula.

Since we found that $k = \frac {3}{160}$ , we can plug it in to the formula, so:

$D = \frac {3}{160} t^2$

Now we can use this to answer all the questions. Since now we have only two unknowns, it means that if we are given one of the unknowns, we can easily solve for the other

**Danielisew** · June 10th 2007, 09:28 AM

oh, ok thanks for explaining it. I understand it better now.

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