Question

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 42 ft/s. Its height in feet after t seconds is given by y=42t−29t2.

Answer 1

ok, and find what ?

Answer 2

A. Find the average velocity for the time period beginning when t=3 and lasting

Answer 3

.005 s

Answer 4

For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator.
B. Estimate the instanteneous velocity when t=3.

Answer 5

do you have an idea how to start ?

Answer 6

yes a little bit. avg=displacement/ time elapsed?

Answer 7

yes, in terms of differentials, we have the same formula as
dy/dt = velocity
where y= displ. , t= time.
so what you get after differentiating y=42t−29t^2.

Answer 8

differentiating y=42t−29t^2. with respect to t

Answer 9

time last? which is 0.05?

Answer 10

differentiate y=42t−29t^2. w.r.t 't'
can you ?

Answer 11

mmm?

Answer 12

ohh!

Answer 13

3+0.05?

Answer 14

\(d/dt (t^n) = n t^ {n-1} \)
so,\( d/dt (y) =d/dt (42t-29t^2)=d/dt(42t)-d/dt(29t^2) \\ =42 d/dt (t)- 29 d/dt (t^2)=...?\)

Answer 15

oh no im confused

Answer 16

say example if y= x^2
dy/dx = 2x right ? (using the formula d/dx (x^n)=n x^{n-1})
now here the variable is 't' instead of x.

Answer 17

\(d/dt (t^n) = n t^ {n-1} \\d/dt (t^2) = ....?\)

Answer 18

nt^2-1!

Answer 19

note that n=2 here.

Answer 20

OK,!

Answer 21

so, \(d/dt (t^n) = n t^ {n-1} \\d/dt (t^2) = 2t \\ d/dt (t)=dt/dt =1\)
got this much ?

Answer 22

doubts ?

Answer 23

yess.. so far.

Answer 24

taking notes please hold

Answer 25

\(d/dt (y) \\ =d/dt (42t-29t^2)\\= d/dt(42t)-d/dt(29t^2) \\ =42 d/dt (t)- 29 d/dt (t^2)\)
look at these steps, and ask if you don't get any particular step...

Answer 26

Have you actually learnt differentiation in class? @Dodo1

Answer 27

\(d/dt (y) \\ =d/dt (42t-29t^2)\\= d/dt(42t)-d/dt(29t^2) \\ =42 d/dt (t)- 29 d/dt (t^2) \\ =42-29 \times (2t) \\ =...?\)

Answer 28

26t?

Answer 29

26? how?

Answer 30

42-29=13. 13*2t. 2t=13 t=13/2

Answer 31

you can't do 42-29
according to PEDMAS, first multiplication needs to be done beffore subtraction.

Answer 32

\(d/dt (y) \\ =d/dt (42t-29t^2)\\= d/dt(42t)-d/dt(29t^2) \\ =42 d/dt (t)- 29 d/dt (t^2) \\ =42-29 \times (2t) \\ so, dy/dt =42-58t\)
make sure you get/understand all steps of above.

Answer 33

It's not possible to teach some how to run when they can't even walk yet. That's something you need to think about @Dodo1

Answer 34

Ok im taking notes please hold

Answer 35

dear whom it may concern, but you can try walk to get to thepoint of how to.

Answer 36

I mean run

Answer 37

Dude, it's a proverb. There's a meaning to that sentence. It means that you have to learn the basics before going ahead.

Answer 38

@Azteck can you please continue your comments in messages ? because they are just a distraction from the problem,Thanks :)
and atleast i am trying...

Answer 39

Thank you for your advice people.. I guess i need to start from basic but I need to finish this problem

Answer 40

so, everything clear till dy/dt = 42 -58t ?

Answer 41

Yes,

Answer 42

I know you are. But it's really difficult to learn maths when your basics are lacking a little. When someone is not understanding fully, you must know when to stop and advise them to learn the stuff that he/she is confused at. No point continuing further, because it's certain that he/she will not understand.

Answer 43

now comes the easy part.
you want average velocity at t=3
so, you need to put t=3 in that equation.

Answer 44

OK so 42-58*3=132

Answer 45

hold on thatws wrong

Answer 46

-48

Answer 47

you missed the - sign again!
42-58*3 = -132

Answer 48

Oh. i should use a cal not on comp

Answer 49

so -132. this is....

Answer 50

no...still few steps are there..

Answer 51

thats dy/dt = -132

Answer 52

Ok!

Answer 53

avg vel.=displacement/ time elapsed
so, you need displacement, and time elapsed given = 0.05
right ?

Answer 54

Yes,

Answer 55

mmm when do I use 3.00?

Answer 56

I mean 0.005

Answer 57

we used dt =0.05
dy = -132 *(0.05)
then, avg. velocity = -132 *0.05 / 0.05 =-132...

Answer 58

mmm so its -132?

Answer 59

i think, yes.

Answer 60

do you have choices/options ?

Answer 61

sorry my email box is not working

Answer 62

No, i dont have any choices

Answer 63

then try entering -132..

Answer 64

it says incorrect

Answer 65

mm:/

Answer 66

i think -132 should work
try -132 ft/s or +132 ft/s

Answer 67

@Azteck any inputs ?

Answer 68

but for 0.01 , the answer says 132.29

Answer 69

-132.29

Answer 70

That equation is pretty much the same as
\[s=u_{y}t+\frac{1}{2}a_{y}t^2\]
What we're finding is the average velocity when t=3 lasting 0.05secs
That 0.05 is putting me off.

Answer 71

average we find in an interval, right... ?
here interval is from 3 to 3.005

Answer 72

You get -135 when t=3 and then I think you sub 0.05 into the equation for t. And then you either subtract or add.

Answer 73

That's what I was going to write.

Answer 74

thank you people sincerely, for helping me tho!

Answer 75

and taking your time!