Question

IS ANYONE HERE TO HELP!!

Answer 1

Do you know what velocity of particle is in moment it reverses direction?

Answer 2

can someone please explain it to me??

Answer 3

@Michele_Laino helloo

Answer 4

I think that the inversion points are those points for which:
\[\frac{ ds }{ dt }=0\]
so first, find the function ds/dt, then find the values of t at which ds/dt=0

Answer 5

can you please do it with me..

Answer 6

@Michele_Laino

Answer 7

3t^2-24t+45=0
t^2-8t+15=0
\[t=\frac{ 8\pm \sqrt{64-60} }{ 2 }=\frac{ 8\pm2 }{ 2 }=3,5\]

find \[\frac{ d^2s }{ dt^2 } at~t=3 and~t=5\]

Answer 8

answer of @surjithayer is right, then write the function \[\frac{ d ^{2}s }{ dt ^{2} }\] @mondona

Answer 9

@Michele_Laino wait what? what do i do next?

Answer 10

please your function, namely s(t) is:
\[s(t)=t ^{3}-12t ^{2}+45t+4\]
so you have to calculate the function ds/dt, namely the derivative over time of s(t), please try to calculate ds/dt

Answer 11

the derivative of that is 3(t^2-8t+15)

Answer 12

right?? can you just write it step by step, i dont really understand your explanation.. @Michele_Laino

Answer 13

that's right, now, in my mind I think that the inversion points are those at which ds/dt=0, so please solve the equation ds/dt=0, using the function you have just found

Answer 14

ok idont know

Answer 15

@Loser66

Answer 16

@surjithayer has made our calculus, you have to solve the equation:
3(t^2-8t+15)=0
try please

Answer 17

3t^2-24t+45=0

Answer 18

yes!

Answer 19

and then?

Answer 20

please solve the quadratic equation above!

Answer 21

when t=3 , 0

Answer 22

noo, please retry

Answer 23

and i got 0 when t=5 also..

Answer 24

ok! Our times are t=3 and t=5, so calculate s(t) at those times, namely insert t=3 and t=5 into the function for s(t)

Answer 25

i got 58 for t=3

Answer 26

right!, now try with t=5 please!

Answer 27

54

Answer 28

Can I say something? @Michele_Laino

Answer 29

then what is next? @Michele_Laino

Answer 30

that's right! you got the answer to the first part of your question. Now to answer to the second part of your question, you have to calculate the second derivative of s(t), namely the function:
\[\frac{ d ^{2} s}{ dt ^{2} }\]

Answer 31

@Loser66 you are welcome!

Answer 32

@mondona you know that s(t) is position function
(s (t))'= v(t) is velocity function
(s(t))'' = a(t) is accelerate function.
right?

Answer 33

58 and 54 are the position?..

Answer 34

So that, the question is about the time the motion "reverse direction" , that is the key.

Answer 35

@mondona that's right, now you have to find the acceleration at t=3 and at t=5

Answer 36

@mondona because you have found them inserting t=3 and t=5 into the function s(t), and the function s(t), from the text of your problem, is the position of your particle

Answer 37

please you have to calculate the function:

Answer 38

\[\frac{ d ^{2} s}{ dt ^{2} }\]

Answer 39

can you show me?

Answer 40

if s'(t)=3t^2-24t+45, then s''(t)=...

Answer 41

6(t-4)

Answer 42

that's right!

Answer 43

now, please insert t=3, and t=5 into the function s''(t), and you will get acceleration at the same times, or the answer to the second part of your question

Answer 44

perfect! ,don't forget to indicate the right measure units, for acceleration and position respectively!

Answer 45

m/sec^2

Answer 46

and position? (:

Answer 47

meters

Answer 48

than you!!

Answer 49

thank you!