# Thread: Help with speed/distance/time and d/dx?

1. ## Help with speed/distance/time and d/dx?

Here is the original problem that I was given:

Carol is a local bicycle racing star and today she is in the race of her life. Moving at a constant velocity k meters per second, she passes a refreshment station. At that instant (t = 0 seconds), her support car starts from the refreshment station to accelerate after her, beginning from a dead stop. Suppose the distance traveled by Carol in t seconds is given by the expression kt and distance traveled by the support car is given by the function $1/3(10t^2-t^3)$ where distance is measured in meters. This latter function is carefully calculated by her crew so that at the instant the car catches up to the racer, they will match speeds. A crew member will hand Carol a cold drink and the car will immediately fall behind.

a. How fast is Carol traveling?
b. How long does it take the support car to catch her?
Originally, I had taken the first derivative of both, set them equal to each other, and solved for t to find the time, then replaced the t in the car's equation with the t that I found. HOWEVER, I took another look at it, and realized that I had differentiated kt as t rather than k. With it being k, I'm honestly not sure how to go about finding the answer to this problem. If I set the first derivative of the car's equation equal to 0, solve for t, replace the t in the car equation with the value I found, solve for distance, and then plug the time and distance into kt and solve for k, then graph both, kt is not actually a tangent line as it should be (which incidentally, differentiating kt as t gave me a tangent line that looked good on the graph - right where it should be) Can anyone help?

2. Originally Posted by Bluekitten
Here is the original problem that I was given:

Originally, I had taken the first derivative of both, set them equal to each other, and solved for t to find the time, then replaced the t in the car's equation with the t that I found. HOWEVER, I took another look at it, and realized that I had differentiated kt as t rather than k. With it being k, I'm honestly not sure how to go about finding the answer to this problem. If I set the first derivative of the car's equation equal to 0, solve for t, replace the t in the car equation with the value I found, solve for distance, and then plug the time and distance into kt and solve for k, then graph both, kt is not actually a tangent line as it should be (which incidentally, differentiating kt as t gave me a tangent line that looked good on the graph - right where it should be) Can anyone help?
It seems to me that at the instant the support car catches up to Carol two things happen:

1. The car and Carol have travelled the same distance, so $\frac{1}{3} (10 t^2 - t^3) = kt$ .... (1)

2. The car and Carol are travelling at the same speed, so $\frac{1}{3} (20 t - 3 t^2) = k$ .... (2)

You need to solve equations (1) and (2) simultaneously for t and k .....

I get t = 5 seconds and k = 25/3 m/s (which is pretty quick! 30 km/hr ........)

3. Originally Posted by mr fantastic
It seems to me that at the instant the support car catches up to Carol two things happen:

1. The car and Carol have travelled the same distance, so $\frac{1}{3} (10 t^2 - t^3) = kt$ .... (1)

2. The car and Carol are travelling at the same speed, so $\frac{1}{3} (20 t - 3 t^2) = k$ .... (2)

You need to solve equations (1) and (2) simultaneously for t and k .....

I get t = 5 seconds and k = 25/3 m/s (which is pretty quick! 30 km/hr ........)
Originally Posted by ninjazking via pm
hi , i have seen u posted an answer for the [above] problem [snip] i would be grateful (sic) if you can give me a short detailed explanation on how did u get T = 5? n K

thank you!
Multiply equation (2) by t and equate the left hand side of the result with the left hand side of equation (1):

$\frac{1}{3} (10 t^2 - t^3) = \frac{t}{3} (20 t - 3 t^2)$.

Expand both sides, simplify and solve for t.

Substitute the value of t into equation (2) and solve for k.