Question

The position of a particle at any time t>=0 is given by x(t)=t^2-2, y(t)=(2/5)t^3. Find the magnitude of the velocity vector at t=4.

Answer 1

I know that the velocity vector is <2t, (6/5)t^2> but how do I find the magnitude of the velocity vector at t=4?

Answer 2

I know the formula for finding the magnitude of the velocity vector at t=4 but the answer is 104/5. How do I simplify sqrt(4t^2+(36/25)t^4) and plug 4 in?

Answer 3

@mathslover

Answer 4

@ybarrap @kirbykirby

Answer 5

\[\sqrt{4t^2+\left(\frac{36}{25}\right)t^4}=\sqrt{t^2\left(4+\left(\frac{36}{25}\right)t^2\right)}=t\sqrt{4+\left(\frac{36}{25}\right)t^2}=\\4\sqrt{4+\left(\frac{36}{25}\right)4^2}=4\sqrt{4+\left(\frac{36}{25}\right)16}=4\sqrt{\frac{100}{25}+\left(\frac{576}{25}\right)}=\\4\sqrt{\frac{676}{25}}=4\frac{\sqrt{4\cdot169}}{5}=4\frac{\sqrt{2^2\cdot13^2}}{5}=4\frac{2\cdot13}{5}=\frac{104}{5}\]

Honestly I don't know what else you can do o.o I mean once you have the formula it is straightforward. But I'm assuming you are not allowed calculators? Because formula can't really be simplified much more the way it is...

Answer 6

on the 2nd line I made the substitution t=4

Answer 7

Is this the only way to solve this problem?

Answer 8

Plug into a calculator :)?

Answer 9

Thanks.

Answer 10

But yeah really there is not much that can be done here... I mean the hard part of the problem was solved by you! Finding 104/5 can be done by calculator, or by hand calculation above... but if you have access to the calculator I highly recommend using it as you save a lot of time.