Particle Movement: Velocity and Acceleration Analysis

[cptstubing]

Homework Statement

A particle moves so that s=108t-t^4, commencing at time t=0 seconds.
a) during what time interval is its velocity positive?
b) during what time interval is its acceleration negative?

Homework Equations

t=0

The Attempt at a Solution

So far, to get velocity (s') I took derivative of above equation, giving me v=108-4t^3
Next, letting v=0 would give me maximum height of particle.
0=108-4t^3
t^3=108/-4
t^3=27
t=3

And so at T=3, velocity is 0, and at t<3, velocity is positive.

Right so far?

Next step, determine when acceleration is negative. First though, can I reason without using any more math that acceleration is negative when t<3? It seems like it would be, at least in a normal ball projectory question, the ball as it approaches it's max height (in this case t=3) it decreases speed at an increasing rate, which is negative acceleration, and when t>3, it is positive acceleration again.

In any case, determining acceleration I took derivative of velocity equation (s')
108-4t^3 ----> -12t^2
Let acceleration = 0
0=-12t^2
12=t^2
3.464=t

So when t<3.464, acceleration is negative, and when t>3.464, acceleration is positive.

My answer for B) doesn't feel right, so I welcome feedback, in fact I really appreciate feedback.

Regards all

 

[SammyS]

Homework Statement

A particle moves so that s=108t-t^4, commencing at time t=0 seconds.
a) during what time interval is its velocity positive?
b) during what time interval is its acceleration negative?

Homework Equations

t=0

The Attempt at a Solution

So far, to get velocity (s') I took derivative of above equation, giving me v=108-4t^3
Next, letting v=0 would give me maximum height of particle.
0=108-4t^3
t^3=108/-4
t^3=27
t=3

And so at T=3, velocity is 0, and at t<3, velocity is positive.

Right so far?

Right so far.

Next step, determine when acceleration is negative. First though, can I reason without using any more math that acceleration is negative when t<3? It seems like it would be, at least in a normal ball projectory question, the ball as it approaches it's max height (in this case t=3) it decreases speed at an increasing rate, which is negative acceleration, and when t>3, it is positive acceleration again.

In any case, determining acceleration I took derivative of velocity equation (s')
108-4t^3 ----> -12t^2
Let acceleration = 0
0=-12t^2
12=t^2
3.464=t

So when t<3.464, acceleration is negative, and when t>3.464, acceleration is positive.

Wrong.

Is t² ever negative?

Then is -12 t² ever positive?

Check your algebra.
0 = -12t² does not give 12 = t² .

My answer for B) doesn't feel right, so I welcome feedback, in fact I really appreciate feedback.

Regards all

 

[cptstubing]

Right so far.

Wrong.

Is t² ever negative?

Then is -12 t² ever positive?

Check your algebra.
0 = -12t² does not give 12 = t² .

Would I take -12t^2 and say:
-12t * (t) = 0

-12t=0 ---> t=0
The whole thing equals zero?

 

[SteamKing]

Homework Statement

A particle moves so that s=108t-t^4, commencing at time t=0 seconds.
a) during what time interval is its velocity positive?
b) during what time interval is its acceleration negative?

Homework Equations

t=0

The Attempt at a Solution

So far, to get velocity (s') I took derivative of above equation, giving me v=108-4t^3
Next, letting v=0 would give me maximum height of particle.
0=108-4t^3
t^3=108/-4
t^3=27
t=3

When you move a number (108) to the opposite side of the = sign, what must you also do?

You've got to be careful. Silly mistakes can lead to wrong answers.

 

[cptstubing]

When you move a number (108) to the opposite side of the = sign, what must you also do?

You've got to be careful. Silly mistakes can lead to wrong answers.

Correct. This works better.
0=108-4t³
-108=-4t³
-108/4=t³
-3=t
Thank you.

 

[SteamKing]

Correct. This works better.
0=108-4t³
-108=-4t³
-108/4=t³
-3=t
Thank you.

No, you're still making silly mistakes with your algebra.

Take the equation:
108 - 4t³ = 0

In order to get an expression containing t on one side, we subtract 108 from both sides of the equation:

108 - 108 - 4t³ = 0 - 108

which leaves

-4t³ = -108

In order to clear the factor of -4 from the LHS, divide both sides of the equation by -4, thus:

-4t³ / -4 = -108 / -4 or

t³ = 27

solving for t by taking the cube root of both sides:

[t³]^1/3 = 27^1/3

therefore, t = 3

In general, unless there are some unusual circumstances, a negative value of t is to be rejected as unrealistic.

You should refresh yourself on basic algebra and arithmetic.

 

[cptstubing]

No, you're still making silly mistakes with your algebra.

Take the equation:
108 - 4t³ = 0

In order to get an expression containing t on one side, we subtract 108 from both sides of the equation:

108 - 108 - 4t³ = 0 - 108

which leaves

-4t³ = -108

In order to clear the factor of -4 from the LHS, divide both sides of the equation by -4, thus:

-4t³ / -4 = -108 / -4 or

t³ = 27

solving for t by taking the cube root of both sides:

[t³]^1/3 = 27^1/3

therefore, t = 3

In general, unless there are some unusual circumstances, a negative value of t is to be rejected as unrealistic.

You should refresh yourself on basic algebra and arithmetic.

Yes I should.
Thanks again.