help with a traveling particle problem please?

[mindshift]

Hello, im taking a practice exam for my AP calculus class, and i wont be able to see the teacher for help on this particular problem. So here it goes:

A particle is traveling along a line so that its position at any time provided that t>=0 is s(t)=\(\displaystyle \int\)6t^2 - 12t dt

at t=1, the particle is 10 units to the right of the origin.

There are several parts to this problem, but i only need help on 2.

1) determine the value of t for which the particle is moving in a positive direction.

2) find the values of t when the particle is slowing down.


I have a feeling that they both require you to take the 2nd derivitive to get the acceleration of the particle, but i dont know where to go from there.

so far from my other problems i have:
s(t) = 2t^3 - 6t^2 + 14
v(t) = 6t^2 - 12t
a(t) = 12t - 12

Any help would be appreciated, thank you.

 

[skeeter]

if \(\displaystyle s(t) = \int 6t^2 -12t dt\), then \(\displaystyle v(t) = s'(t) = 6t^2 - 12t\) and \(\displaystyle a(t) = v'(t) = 12t - 12\)

the sign of velocity, \(\displaystyle v(t)\), tell you the direction of travel.

the particle will be slowing down when \(\displaystyle v(t)\) and \(\displaystyle a(t)\) have opposite signs.

 

[galactus]

If s'(t)=v(t)>0 at a given time t, then the particle is moving in the positive direction.

The particles velocity is decreasing when s''(t)=a(t) is <0

 

[skeeter]

fyi ... a decrease in velocity does not mean the object in question is slowing down.

 

[mindshift]

fyi ... a decrease in velocity does not mean the object in question is slowing down.

are you talking about a change in direction?

i have the position, velocity, and acceleration formulas. So what you both are saying is that i need to show that v(t) is moving in the positive direction when t is greater than 0, and that velocity is decreasing when a(t) is less than 0 because its a negative acceleration, correct?

So, to show this i would simply state these two facts as:

the velocity of the particle must be greater than 0 [v(t)>0] in order for it to travel in a positive direction.

the particle is slowing down when its acceleration is less than 0, a.k.a. a negative aceleration [a(t)<0]


is there any mathematical work i can do to back this up, or should this be fine?

 

[galactus]

I'm sorry, I should leave the physics to you.

Here's part of what I always thought. Correct me if I'm wrong:

If s(t)>0, the tangent has positive slope and the curve is concave down,
then particle is on the positive side of the origin, particle in moving in positive direction, velocity is decreasing and particle is slowing down.


If s(t)>0, tangent line has negative slope and curve is concave down, then
particle is on positive side of origin, particle is moving in negative direction, velocity is decreasing and particle is speeding up.


If s(t)<0, tangent line is negative, concave up, then particle is in negative position, particle is moving in negative direction, velocity is decreasing and particle is slowing down.


If s(t)>0, tangent is 0, curve is concave down, then particle is in positive position, is stopped for time being and velocity is decreasing.

I'll leave the physics to you from now on.

 

[skeeter]

fyi ... a decrease in velocity does not mean the object in question is slowing down.

are you talking about a change in direction?

i have the position, velocity, and acceleration formulas. So what you both are saying is that i need to show that v(t) is moving in the positive direction when t is greater than 0, and that velocity is decreasing when a(t) is less than 0 because its a negative acceleration, correct?

So, to show this i would simply state these two facts as:

the velocity of the particle must be greater than 0 [v(t)>0] in order for it to travel in a positive direction. yes

the particle is slowing down when its acceleration is less than 0, a.k.a. a negative aceleration [a(t)<0] no ... as I stated before, the particle is slowing down when acceleration and velocity have opposite signs


is there any mathematical work i can do to back this up, or should this be fine?

first of all, the domain for time in this problem is stated at t > 0.

v(t) = 6t² - 12t = 6t(t - 2)
v(t) < 0 on the interval 0 < t < 2 ... the particle moves left
v(t) > 0 on the interval t > 2 ... the particle moves right
v(t) = 0 at t = 0 and t = 2 ... the particle is at rest

a(t) = 12t - 12 = 12(t - 1)
a(t) < 0 on the interval 0 < t < 1
a(t) > 0 on the interval t > 1
a(t) = 0 at t = 1.

now, you figure out this part ... v(t) and a(t) have opposite signs where?

 

[pka]

I will hop into this.
Look at this comparsion.

 

[mindshift]

ok, from looking at the graph, v(t) and a(t) have opposite signs from x=1 to x=2, so from that interval the particle is slowing down.

 

[soccerball3211]

When i graphed the velocity of the particle i found that the sign of the velocity is positive when t>1. So is this when the particle is moving in a positive direction?

 

[skeeter]

When i graphed the velocity of the particle i found that the sign of the velocity is positive when t>1. So is this when the particle is moving in a positive direction?

if v(t) = 6t² - 12t, then v(t) > 0 for t > 2.

look at pka's graph in blue.