How do I solve this linear equation using constant acceleration equations?

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Homework Help Overview

The discussion revolves around solving a linear equation related to constant acceleration in a physics context. The original poster is attempting to find the time at which a particle, initially moving vertically and accelerating horizontally, reaches a specific angle with respect to the horizontal.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the original poster's approach to solving for time using a tangent function and constant acceleration equations. There are questions about the validity of the calculations and the setup of the problem, particularly regarding the signs of the velocities and accelerations. Some participants suggest breaking the problem into components and clarifying the initial conditions.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem. There is a focus on clarifying assumptions and ensuring the correct interpretation of the physics involved. Multiple interpretations of the problem setup are being explored, and no consensus has been reached yet.

Contextual Notes

Participants note the importance of the angle being in degrees and question the initial conditions of the particle's motion. There is uncertainty regarding the signs of the initial velocity and acceleration components.

Osbourne_Cox
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1. I thought I was doing it right, but Quest said I have to wrong answer. Can some one solve it in steps and produce a final answer so I can compare?

Tan 58=(vi-ayt)/axt

where vi=5.6
ay=-9.8
ax=2.1

solve for t.

I got 0.87 for t.




Thank you.
 
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Could you please include your attempt to a solution?
 
vi-at=tan 58 (at)
-(-9.8)t=tan58(2.1)t-5.6
9.8t=3.36t-5.6
6.44t=-5.6
t=0.87
 
Take a look at the second to last step. You have:

6.44*t = -5.6

Your answer should be negative.
 
How do we have negative time though? (its the math to a physics problem)
 
Does your problem require you to be in degrees or radians?
 
Degrees, and my calculator was set on degrees.
 
Maybe you should post the original physics problem. Is it possible you made an error setting the problem up :o?
 
Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?
 
  • #10
Sorry, I was not aware that 't' is time.
 
  • #11
Oh, it is very possible, ha ha. I am not good at this stuff at all.
 
  • #12
Osbourne_Cox said:
Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?

At t = 0, is the thing moving up or down? If it's moving up, vi will have a sign opposite to that of gravitational acceleration. Your equation in the first post suggests that the thing is moving up.
 
  • #13
I assume you are using equations for constant acceleration. You are going to need to break your equations into components: one for your vertical velocity and acceleration components and one for your horizontal components.

Vy=Viy+ay*t
and
Vx=Vix+ax*t

Where
Vy=V*sin(58)
Vx=V*cos(58)

You end up with two equations and two unknowns (V and t). You should be able to solve.
 

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