Question

the position of a particle as it moves along the x-axis is given for t>0 byt x=(t^3-3t^2+6t) [m] where t is in [s]. Where is the particle when it acheives minimum speed after t=0?

Answer 1

If you know the position of the particle x(t), how might you try to find its speed v(t) ?

Answer 2

1st derivative

Answer 3

Yes, calculate dx/dt

Answer 4

ok thats easy, 3t^2-6t+6

Answer 5

but this is just telling me the slope at a position and does not specify a position

Answer 6

ok, and the question asks about the minimum speed of the particle
so if you know v(t), what do you need to do to locate its minimum value ?

Answer 7

we have to get to the answer in a few steps you see

Answer 8

to see if there are any zeros or minimums?

Answer 9

yea i see thats why i am having a hard time understanding cuz im trying to figure out these relationships

Answer 10

minima, we want to find where, or more precisely when, v is a minimum

Answer 11

yes

Answer 12

so then we could find the derivative of that to get the acceleration which would be the slope of v(x)

Answer 13

so you understand that when you differentiated x(t), the result you got is actually v(t) ?

Answer 14

v(t)*

Answer 15

yes

Answer 16

the instantaneous velocity/speed

Answer 17

ok, and you are right , if you differentiate v(t) that will tell you the acceleration a(t)

Answer 18

which is the slope of the v(t) graph

Answer 19

correct
and what value does the slope of a graph have when the graph is at a minimum ?

Answer 20

0?

Answer 21

yes

Answer 22

so i need to find the second derivative and where is equals 0

Answer 23

and that would give me a specific point

Answer 24

so, if you find the time when acceleration is 0, you know when the velocity is at a minimum

Answer 25

differentiate v to find a, tell me what you get for a(t)

Answer 26

that point is where acceleration is 0 which would be the minimum

Answer 27

ok so a(t)=6t-6

Answer 28

correct
so you can see what value of t gives a=0 ?

Answer 29

yes t=1, which would be the point where a(t)=0 which would mean minimum v(t)

Answer 30

that's right
actually v might be a maximum, but it is a minimum in this case

Answer 31

so how can you finally answer the question and figure out where the particle is, when v is a minimum

Answer 32

so i can take the t value and plug it in the original equation and that would tell me the x value at that point

Answer 33

you got it !

Answer 34

What do you get for x at 1 sec ?

Answer 35

x=1 at 1 sec

Answer 36

x=4

Answer 37

sorry hit wrong key

Answer 38

: )

Answer 39

problem solved
if you wanted to be really thorough you could look at how v(t) behaves as you move away from t=1, to check that it really is a minimum and not a maximum

Answer 40

or you can check it by looking at how a(t) behaves either side of the minimum/maximum, either way

Answer 41

yea i see, at first i was trying to graph the positions in relation to time and i knew the derivative was slope of V(t) but didnt know where to take it to find the actual point

Answer 42

i appreciate it very much, i just started this physics class and am trying to grasp these concepts, was going to go to a tutor the end of the week but found out they were only there mon-wed; i did all the reading yesterday and thought i understood

Answer 43

i know it can be tough sometimes, hope it seems a bit clearer now

Answer 44

yea that one is

Answer 45

now i am trying to figure out one that give me the velocity equation but wants the accel. when it achieves max displacement in positive x-direction

Answer 46

so do i have to do the integral to find the equation that would show displacement

Answer 47

you might not need to find x

Answer 48

if you think about it, when the displacement is a maximum, what will the value of v be ?

Answer 49

lowest

Answer 50

no it would be highest because is is the numerator?

Answer 51

v is the derivative of x, and x is at a maximum, so v will have a very particular value

Answer 52

it's the same kind of idea we had in the first problem, when a function is at a maximum or minimum, what is the value of its derivative ?

Answer 53

0

Answer 54

correct

Answer 55

so, if x(t) is at a max or min, what will the value of v=dx/dt be ?

Answer 56

so i can take the derivative of the V(t) to get a(t), the take the derivative of that and set is equal to 0

Answer 57

no, slow down

Answer 58

think about x(t), it's just some function, right ?

Answer 59

yes

Answer 60

if x(t) is at a min or max then V(t) would equal 0

Answer 61

yes, that's right

Answer 62

and at that point accelleration would equal 0

Answer 63

no reason why it should

Answer 64

find the value of t when v=0

Answer 65

then use that value of t to calculate a(t) and see what comes out, it doesn't have to be zero

Answer 66

so when i set Vx=0 in the original problem is almost impossible to solve because its Vx=(32t-2t^3)

Answer 67

couldnt i take the derivative up until i can get a zero