Question

The position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

Answer 1

@ganeshie8 @Hero @One098 @saifoo.khan @eliassaab @abb0t @PaxPolaris

Answer 2

so if the equation was written

y= -6t-2

can you frind the derivative dy/dt?

Answer 3

how would I use that equation?

Answer 4

the equation is exactly the same as your equation - I have just written it in a way that might be more familiar to you

we have a fiunction s(t) . I have just said let y=s(t) (so we write y in place of s(t)

I then just re-arranges the -2-6t into -6t-2 (because we normally put equations in descending order of the variable.

so we have

displacemet y = -6t-2

can you write the equation for the velocity? (=dy/dt)

Answer 5

I forgot how to...like which numbers go in for d x and y?

Answer 6

I guess that wasn't helpful

going back to the original question you are given

s(t) = -2 - 6t

can you write the derivative ds/dt

Answer 7

err could you show me?:) im a bit lost

Answer 8

no sorry - you are evidently studying derivitives. d is not a number - it is a notation to show that you differentiate the equation.

If oyu do not know how to differentiate -6t-2 then you need to go back to your text, or your teacher, since this is th efirst step in a LOOOOONG road of calculus and you need to know this step before you can move on

Answer 9

lim x->2 = f(x+h) - f(x) / h
would I start like this tho? like could this work?

Answer 10

The formula for velocity given constant acceleration is v = u + at, where u is the velocity when the time, t, is equal to 0 and a is the rate of acceleration. The acceleration can be expressed as the change in velocity over the change in time. The instantaneous velocity can be found by taking the derivative of the equation in respect to distance. In this case, velocity and speed are the same, except that velocity is expressed as vector and speed is expressed in scalar form.

Answer 11

sure you can use t=2 .... but the limit is alwayys: h -> 0

Answer 12

\[\large \lim_{h \to 0} {s(2+h) - s(2) \over h}\]

Answer 13

well how would I use that to solve?

Answer 14

well what is s(2)?

Answer 15

-14?

Answer 16

right .... and what's s(2+h)

Answer 17

how do I know what h is?

Answer 18

we don't need a number for h

\[s(2+h) = -2-6(2+h) \ ...\]

Answer 19

\[= -2 -12 -6h\\ =\color{blue}{-14-6h}\]

Answer 20

just plug this in you derivative equation .... the h will cancel out and you get the velocity

Answer 21

so would it be (-14-6h)x/(-14-6h)y ?

Answer 22

the derivative at t=2\[= s'(2)\\ =\lim_{h \to 0} {s(2+h)-s(2) \over h}\\ = \lim_{h \to 0} {\left[ -14-6h \right]-\left[ -14 \right] \over h}\\=\ ...\]

Answer 23

so it would end up being -6 right?

Answer 24

@babygirl202
sorry if I was a little short above - I had assumed that you were a bit further along the road of working out derivitives.

The route above is quite a long-winded way to get to the answer - but it will get you there.

Answer 25

\[=\lim_{h \to 0} {\cancel{-14}-6h+\cancel{14} \over h}\\ =\lim_{h \to 0} {-6h \over h}\\=\lim_{h \to 0} {-6 \cancel h \over \cancel h}\\=\lim_{h \to 0} -6\\=-6\]

Answer 26

yep you got it :)

Answer 27

you can see that th evelocity is a constant (-6) and hence does not depend on t

Answer 28

so then would that mean the instantaneous velocity at t=2 is -6?

Answer 29

yes! because velocity is the derivative of the position function (with respect to time).

Answer 30

the original expression was s= kt+b

that is th eequation for a straight line

the SLOPE of the line is the velocity, and in this case the slope is k

in all cases whwere you have a linear equiation (y=kt) the derivative will be k at all values of t

Answer 31

alright thank you guys so much for your help:)