Velocity and Acceleration homework

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SUMMARY

The discussion focuses on calculating the velocity of point A given its acceleration defined by a=-1.8sin(3t) m/s², with initial conditions x=0 and v=0.6 m/s at t=0. The user integrated the acceleration equation to find the velocity function, resulting in v=0.6cos(3t) + v0. Upon substituting t=0.5s, the calculated velocity was 0.6424 m/s, which contradicts the textbook answer of 42.4 mm/s. The discrepancy arises from the misunderstanding of the initial velocity's role in the integration process.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with kinematics equations, particularly acceleration and velocity relationships.
  • Knowledge of trigonometric functions and their properties.
  • Basic understanding of units conversion, especially between meters per second and millimeters per second.
NEXT STEPS
  • Review integration techniques for solving differential equations in physics.
  • Study kinematic equations and their applications in motion analysis.
  • Learn about unit conversions, particularly between m/s and mm/s.
  • Explore the role of initial conditions in solving differential equations.
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Students studying physics, particularly those focusing on mechanics and kinematics, as well as educators looking for examples of solving motion problems involving acceleration and velocity.

srh
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Homework Statement


The acceleration of point A is defined by the relation a=-1.8sin(kt), where a and t are expressed in m/s^2 and seconds, respectively, and k=3 rad/s. Knowing that x=0 and v=0.6m/s when t=0, determine the velocity of point A when t=0.5s.


Homework Equations


a=dv/dt
a dt = dv
a=-1.8sin(3t)

The Attempt at a Solution


I began by using the equation a dt=dv. I substituted -1.8sin(3t) in for a. After integrating both sides I got v-v0=0.6cos(3t). I plugged in 0.6 m/s for v0. For t=0.5 I got v=.6424 m/s. The answer in the back of th book was 42.4mm/s. If I left v0 as zero, I would get that answer. But I thought I needed to include that. Am I missing something?
 
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srh said:

Homework Statement


The acceleration of point A is defined by the relation a=-1.8sin(kt), where a and t are expressed in m/s^2 and seconds, respectively, and k=3 rad/s. Knowing that x=0 and v=0.6m/s when t=0, determine the velocity of point A when t=0.5s.


Homework Equations


a=dv/dt
a dt = dv
a=-1.8sin(3t)

The Attempt at a Solution


I began by using the equation a dt=dv. I substituted -1.8sin(3t) in for a. After integrating both sides I got v-v0=0.6cos(3t). I plugged in 0.6 m/s for v0. For t=0.5 I got v=.6424 m/s. The answer in the back of th book was 42.4mm/s. If I left v0 as zero, I would get that answer. But I thought I needed to include that. Am I missing something?

Think about the highlighted text. What function of the form v = f(t) is going to give you v0=0.6m/s?
 

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