Particle Movement: Velocity and Acceleration Analysis

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In summary, at time t=3 the particle's velocity was 0 and its acceleration was either negative or positive depending on how close to 3 it was.
  • #1
cptstubing
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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
 
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  • #2
cptstubing said:

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
 
  • #3
SammyS said:
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?
 
  • #4
cptstubing said:

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.
 
  • #5
SteamKing said:
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.
 
  • #6
cptstubing said:
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 = 271/3

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.
 
  • #7
SteamKing said:
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 = 271/3

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.
 

1. What is a moving particle question?

A moving particle question is a problem or experiment that involves the motion of a single particle or object. It often requires the use of mathematical equations and principles of physics to analyze and solve.

2. How do you calculate the velocity of a moving particle?

The velocity of a moving particle can be calculated using the equation v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken to travel that distance. The units for velocity are typically meters per second (m/s).

3. What is the difference between speed and velocity?

Speed refers to the rate at which an object is moving, while velocity refers to the rate at which an object is changing its position. Velocity takes into account both the speed and direction of the object's motion, whereas speed does not.

4. How does acceleration affect the motion of a particle?

Acceleration is the rate of change of velocity over time. This means that an object with a higher acceleration will experience a greater change in velocity over a given period of time. Therefore, acceleration plays a significant role in determining the motion of a particle.

5. What are some real-life applications of moving particle questions?

Moving particle questions have many real-life applications, such as predicting the trajectory of a thrown ball, understanding the motion of celestial bodies, and designing vehicles or machines that require precise movements. They are also used in fields such as engineering, physics, and robotics.

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