Question

Can someone help me figure out the third part of this problem? I got part a to be -38.4 m/s^2 and part b to be 0.52 seconds, but I dont know how to do the part c. Please see the attached image for the problem.

Answer 1

Can you write the magnitude of the velocity?

Answer 2

where do you want me to write it?

Answer 3

Anywhere you want.

Answer 4

ok, do you mean, v = (5.2t - 5.0 t^2) i + 8.7 j?

Answer 5

What is the magnitude of that vector?

Answer 6

I dont really know what you want me to say? 10 m/s?

Answer 7

im a bit confused

Answer 8

sorry I got disconnected..

Answer 9

yea me too

Answer 10

By writing the magnitude of the velocity I meant finding
\[V = \sqrt{V _{x}^{2} + V _{y}^{2}}\]

which is the magnitude of the vector V with x component Vx and y component Vy

Answer 11

ok, I think I see what you mean. So for Vx it would be (5.2t - 5.0t^2) and for Vy it would be (8.7 j)...then square those to plug in the formula? or is that way off?

Answer 12

plug em in and set V = 10

solve for t

Answer 13

ok

Answer 14

I plugged it all in and expanded to get Sqrt.(27.04t^2 - 52t^3+25 t^3+ 75.69)...how do I reduce that?

Answer 15

Pretty messy huh?

Answer 16

yea :)

Answer 17

is there a way to get rid of something so that i can use the quadratic formula?

Answer 18

I'm working on it.

Answer 19

ok

Answer 20

I got
\[\left( 5.2t - 5t ^{2} \right)^{2} = 24.3\]

Do you see how I got it?

Answer 21

did u factor out a t^2?

Answer 22

Not really, I didn't expand the expression just collected the numbers

Answer 23

like used the quadratic formula on the inside values of the paranthesis without expanding
?

Answer 24

yes.I started with
\[100 = 8.7^{2} + \left( 5.2t + 5t ^{2} \right)^{2}\]

Answer 25

I get 1.04 = t? did u get that too?

Answer 26

Yep High Five!!

Answer 27

But then how did you get the 24.3 number? Cause I tried plugging it in and for that part I get more like 112.5?

Answer 28

100 - 8.7*8.7 = 24.31

Answer 29

Oh, I thought that if you plugged t = 1.04 s into \[(5.2 t + 5.0t^2)^2\], you would get 24.31

Answer 30

your coefficient of T^2 should be negative..

Answer 31

oh, right, I must have added a + sign for no reason :) always the little mistakes that get me...So the answer would be 1.04 s?

Answer 32

I'm currently in a mental fog. Can't see to get a reasonable answer hold on

Answer 33

OK found the problem, messy paper, got= 1.64 sec

Answer 34

Sorry it took so long.lesson be neat.

Answer 35

in the quadratic formula? how did you get that? because I got 1.04 too?

Answer 36

I' m solving
\[t \left( 5.2 - 5t \right) = \pm 4.93\]

Answer 37

if you set 5.2 - 5t = 0 and solve for t, you still- get 1.04, right?

Answer 38

Can't do that. It's not a solution to the equation if 5.2-5t =0 then you have 4.93 =0

Answer 39

im not following..where do you get that number?

Answer 40

\[t \left( 5.2 - 5t \right) = \pm 4.93 \] is the equation we are solving. t cannot equal 0 and 5.2-5t cannot equal 0 otherwise the equation make no sense.

Answer 41

oh, ok, I think I get where you are coming from...u did the square root of 24.31 and set that equal to t(5.2 - 5t)...then u solved for t to get 1.64?

Answer 42

Yes

Answer 43

did you set it to 0 and use the quadratic formula?

Answer 44

yes is solved
\[5t ^{2}- 5.2t \pm4.93 = 0\]

Answer 45

Ok, that sounds good

Answer 46

thank you so much for spending the whole time helping me out!! :) that was really nice

Answer 47

you welcome and good luck with your studies