Question

Suppose that the position of a particle as a function of time (in seconds)is given by the formula s(t)=5.5+5t^3-t^5, t>0

a)The time at which the velocity is zero is
b)The time at which the acceleration is zero is
c)The time at which the velocity is maximum is

Answer 1

velocity is ds/dt,time at which the velocity is zero is the time at which ds/dt=0

Answer 2

can you show me how to do it, im confused

Answer 3

ds/dt=5.5+15t^2-5t^4 is that understood?

Answer 4

sorry ds/dt=15t^2-5t^4

Answer 5

but you need to understand why

Answer 6

yea its the derrivitve

Answer 7

okay ..then we need 15t^2-5t^4=0

Answer 8

so itll be 5t^2(3-t^2)=0

Answer 9

or better 5(3t^2-t^4)=0

Answer 10

and then because 5 cant be zero we need 3t^2-t^4 to be zero

Answer 11

so 3t^2-t^4=0 ----->3t^2=t^4 ok till now?

Answer 12

yes

Answer 13

3t^2=t^4 ----->3=t^2

Answer 14

yes

Answer 15

so \[t=\sqrt{3}\] or \[t=-\sqrt{3}\]

Answer 16

ok till now?

Answer 17

yese

Answer 18

can we have negative time? :o

Answer 19

no

Answer 20

so its sqrt(3)

Answer 21

then we have \[t=\sqrt{3}\] seconds

Answer 22

thanks... for the acc. do we set the second derr. to zero?

Answer 23

yes

Answer 24

30t-20t^3=0
10(3-2t^2)=0

Answer 25

3-2t^2=0
3=2t^2

ans sqrt(3/2)

Answer 26

how do we find part c?

Answer 27

good job on part 2

Answer 28

how do we find part c?

Answer 29

umax=at right?

Answer 30

?

Answer 31

the maximum speed is acceleration*time

Answer 32

yes

Answer 33

so what do we do next

Answer 34

accelaration*time=30-20t^3*t and velocity=15t^2-5t^4

Answer 35

sorry acceleration*time=(30-20t^3)*t

Answer 36

damn (30t-20t^3)*t

Answer 37

okay

Answer 38

30t^2-20t^4=15t^2-5t^4

Answer 39

okay

Answer 40

then we divide right and left hand side with t^2

Answer 41

Hmmm? Use the velocity equation to find the velocity at a particular time.

Answer 42

okay

Answer 43

Use acceleration equation to find critical points in time.

Answer 44

max and min are at critical points.

Answer 45

yes but because u=at we have u and a so dont we find t from there?

Answer 46

The purpose of b) was to find \(t\).

Answer 47

oh damn yeah my bad i see what you mean

Answer 48

It's an open interval so there is only one critical point and if that isn't the maximum then there is no maximum. Open intervals are not subject to extreme value theorem.

Answer 49

so what do we do?

Answer 50

Plug in \(t\) from b) into your velocity equation.

Answer 51

you need to put sqrt(3/2) into 5(3t^2-t^4)

Answer 52

and then set it equalto zero
?

Answer 53

no you ll find a number..that's the maximum velocity

Answer 54

wait im confused

Answer 55

we multpile them?

Answer 56

you put where t is at the velocity equation the acceleration we found at question b that is sqrt(3/2)

Answer 57

so maximum speed=15*sqrt(3/2)^2-5sqrt(3/2)^4

Answer 58

you put the result of our b question=acceleration into our velocity equation we got from question a

Answer 59

what answer are you getting as the final, its saying my answer is wrong

Answer 60

yes i see why

Answer 61

i got a bit confused with question c...brb checking my old kinematic notes

Answer 62

k

Answer 63

still seems the procedure is right...what result did u find?

Answer 64

i just entered it exactly like that before the final answer and it says inccorect

Answer 65

what are you getting?

Answer 66

the final answer is 11.43?

Answer 67

no its coming up wrong

Answer 68

part b is correct im just having a problem with part c

Answer 69

yes it is i dont know i still get 11.43

Answer 70

whats the whole number you got... any numbers after .43?

Answer 71

basically its 11,271

Answer 72

i redid it

Answer 73

so would be 11.3

Answer 74

its still coming up wrong.. ill just leave that for now

Answer 75

thanks so much though!

Answer 76

how much is the answer?

Answer 77

ah wait

Answer 78

yea

Answer 79

i dont know the answer yet, did you find it?

Answer 80

we found the max velocity,not the time

Answer 81

(30t-20t^3)t=11.3

Answer 82

okay

Answer 83

now what?

Answer 84

30t^2-20t^4=11.3

Answer 85

okay

Answer 86

i find imaginary parts in result ,must be wrong,sec

Answer 87

k

Answer 88

seems like the velocity is at a maximum the time acceleration is zero that means t=3/2=1.5 sec