Question

Find the equation for a linear trajectory that starts at (0,0) and ends at (300,400) where the speed increases according to S(t) = 2t. Begin by assuming the equation of trajectory has the form r(t) = where u’(t) > 0.

Answer 1

Your going to want to use the Pythagorean theorem and maybe some trigonometry to get its position in the horizontal and vertical directions as a function of t, I am to lazy to work out the details.

Answer 2

@Charollete, did you understand my approach for your other question like this one where speed was constant 25 ?

Answer 3

No. but for this one the speed isn't constant, but for the integral of 2t it would be t^2. The last part is what's throwing me off for this one.

Answer 4

Also made a mistake in the question

Answer 5

Would it be the same approach x(t) = t^2 and y(t) = t^2 since the start is at <0,0>

Answer 6

@dumbcow Not sure of your approach, I only saw your answer.

Answer 7

particle travels 500 units
\[r(t) = t^{2} = 500\]
solve for t
\[t = \sqrt{500} = 10\sqrt{5}\]
\[x(t) = bt^{2} \]
\[500b = 300\]
\[b = \frac{3}{5}\]
\[x(t) = \frac{3}{5}t^{2}\]
similarly y(t) can be found just set it equal to 400
\[y(t) = \frac{4}{5}t^{2}\]

Answer 8

in component form:
\[r(t) = <\frac{3}{5}t^{2}, \frac{4}{5}t^{2}>\]

Answer 9

@dumbcow so for the other problem I did would it be wrong in comparison to this one? I was shown a different way and the t values weren't matching up when I solved for them.

Answer 10

Also @dumbcow how would I go about finding how much time for the object or is that just what you found earlier \[10\sqrt{5}\] ?

Answer 11

At the end point that is.

Answer 12

@Charollete , yes when you solve for "t" that is how much time it takes to get from one end point to the other

for other problem it can be done same way...constant speed of 25
\[r(t) = 25t\]
the distance from (0,50) to (200,350) is 100sqrt13
\[25t = 100\sqrt{13}\]
\[t = 4\sqrt{13}\]
\[x(t) = bt = 200\]
\[4\sqrt{13}b = 200\]
\[b = \frac{50}{\sqrt{13}}\]
\[x(t) = \frac{50}{\sqrt{13}}t\]
then for y(t) set endpoint equal to 350, note initial value of 50
\[ct +50 = 350\]
\[4\sqrt{13} c = 300\]
\[c = \frac{75}{\sqrt{13}}\]
\[y(t) = \frac{75}{\sqrt{13}}t + 50\]

Answer 13

@dumbcow that makes a lot of sense, was the other way wrong, with how she was showing me to do it?

Answer 14

i dont think it was wrong...just didn't consider the total distance traveled as determining the time interval

Answer 15

Ahh. alright that makes sense, thank you!

Answer 16

no problem