Question

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Answer 1

@hartnn sambhalo plz

Answer 2

he will be able to help u better..:)

Answer 3

isnt velocity the derivative of position? v = s'

Answer 4

then the diffy q is can be written as

s' = vo - ks

s' + ks = vo and find solutions

Answer 5

your work looks fine, now recall that s(0) = 0, so when t=0, s=0

Answer 6

e^(a+b) = e^a e^b

Answer 7

since c is arbitrary then e^(kc) is arbitrary as well

vo - ks = Ce^(kt)

Answer 8

now

s' = vo - ks

s' = Ce^(kt)

calculate for s

Answer 9

s' = Ce^(-kt) ... :)

Answer 10

the thing is C is an arbitrary constant, its can be this or that or something else which makes e^kc an arbitrary constant in and of itself right?

Answer 11

what is the value of e^kc when c can be any value it wants to be?

Answer 12

why do u need to separate the constant ?

Answer 13

i dont see why

s' = vo - ks = e^(kc-kt) = e^(kc) e^(-kt)

s' = Ce^(-kt)

Answer 14

\(s = \dfrac{v_0 }{k}- Ce^{-kt} \)

is the best you can simplify

Answer 15

@hartnn ok ..that's is what i wanted.. to simplify my answer

Answer 16

also for part b do we only differentiate v=v_0 -ks to get the eq for acceleration?

Answer 17

silly me, i was gonna work from s' into that :) solving for s outright seemed to simple lol

Answer 18

well, since s' = Ce^(-kt), acceleration is s''

Answer 19

v' would be easier :P

Answer 20

only after you plug in the function of s in terms of t in v

Answer 21

v' = -ks' ;) good luck