Velocity of Particle Homework: Find Acc, Zero Accel, & Speed=10 m/s

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Homework Help Overview

The problem involves a particle's velocity in the xy plane, expressed as a function of time, with specific questions regarding acceleration, conditions for zero acceleration, and instances when speed equals 10 m/s.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between velocity and speed, questioning how to set up equations involving time and its squared term. There is confusion about manipulating the velocity equation to find speed and how to isolate terms.

Discussion Status

Some participants have provided attempts at solutions for parts of the problem, while others express confusion about the setup of equations and the implications of the terms involved. There is ongoing exploration of how to approach the calculation for speed.

Contextual Notes

Participants are grappling with the mathematical representation of velocity and speed, particularly in relation to the time variable and its squared form. There is a lack of consensus on how to proceed with the calculations for speed.

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Homework Statement


The velocity v of a particle moving in the xy plane is given by v= (5.9 t - 4.1 t2)i + 8.7j, with v in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.7 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?


Homework Equations





The Attempt at a Solution


Part A is -24.44 i + 0 j m/s2

Part B is .7195 seconds.

How do I find part C?
 
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if velocity is v=xi+yj

then |v| =speed = √(x2+y2)
 
That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?
 
tjbateh said:
That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?

the 'y' term is a constant, so you can square both sides of the equation and then make the 'x2' the subject and then take the square root of both sides, then solve. (I hope you understood what I meant)
 
No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??
 
tjbateh said:
No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??

ok we'd get


[tex]10=\sqrt{(5.9t-4.1t^2)^2 +(8.7)^2}[/tex]


so if you square both sides you get rid of the square root sign. Then rearrange and make (5.9t-4.1t^2)2 the subject and take the square root of both sides now.
 
would I get sometime like 1.3= t(5.9-4.1t)??
 

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