Kinematics moving particle problem

In summary, the position of a particle moving along the x-axis is described by the equation x=ct^{2}-bt^{3}, where x is in meters and t is in seconds. The units of the constants c and b are meters per seconds squared and meters per seconds cubed, respectively. The particle reaches its maximum positive x position at t=1 second. From t=0 to t=2.0 seconds, the particle moves a distance of 2 meters and has a displacement of 4 meters. Its velocity at times 1.0, 2.0, 3.0, and 4.0 seconds is 2 m/s, 4 m/s, 6 m/s, and 8
  • #1
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Homework Statement


The position of a particle moving along the x-axis depends on the time according to the equation

x=ct[tex]^{2}[/tex]-bt[tex]^{3}[/tex]

where x is in meters and t in seconds. What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0 respectively.(c) At what time does the particle reach its maximum positive x position? From t=0 to t=2.0, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 (g) 2.0 (h) 3.0 and (i)4.0. Find its acceleration at times (j) 1.0 (k) 2.0 (l) 3.0 and (m)4.0.

Homework Equations



x=ct[tex]^{2}[/tex]-bt[tex]^{3}[/tex]

The Attempt at a Solution



I can't even get started on part a. I am not sure how to find the constants. The only thing I could think of would be using the final unit (m), so in that case one would have to be meters and the other seconds?
Am I on the right track?
 
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  • #2
You have x=ct2-bt3

'x' is in meters(m), so that the quantities ct2 and bt3 must also have units meters(m).

You know that 't' is in seconds (s), so 't2' has units s2, so what should the units of 'c' be such that (units of c)*(units of t2)= meters?
 
  • #3
OK so for part A I have c=[tex]\frac{m}{s^{2}}[/tex] and B as b=[tex]\frac{m}{s^{3}}[/tex]. I believe the unit s should cancel out.
On part c I need to find a relative maximum. I can look at the graph and tell that t should equal 1. I was thinking to use the formula x=[tex]\frac{-b}{2a}[/tex] then plug x in for t in the given equation, but I don't think I can use that because it is not a quadratic.
 

Related to Kinematics moving particle problem

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What is a moving particle problem?

A moving particle problem is a type of kinematics problem that involves analyzing the motion of an object or particle in terms of its position, velocity, and acceleration over time.

3. How do you solve a kinematics moving particle problem?

To solve a kinematics moving particle problem, you need to identify the given information, such as initial and final positions, velocities, and accelerations. Then, use the kinematic equations to find the unknown quantities.

4. What are the three kinematic equations?

The three kinematic equations are:

1. v = u + at (equation for final velocity)

2. s = ut + 0.5at^2 (equation for displacement)

3. v^2 = u^2 + 2as (equation for final velocity squared)

5. What are some common types of kinematics moving particle problems?

Some common types of kinematics moving particle problems include: projectile motion (objects moving through the air under the influence of gravity), circular motion (objects moving in a circular path), and simple harmonic motion (objects moving back and forth in a periodic motion).

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