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

AI Thread Summary
To solve the linear equation using constant acceleration, the initial vertical velocity (vi) is 5.6 m/s, with vertical acceleration (ay) at -9.8 m/s² and horizontal acceleration (ax) at 2.1 m/s². The goal is to determine the time (t) when the particle travels at an angle of 58° to the horizontal. The discussion highlights the need to separate the vertical and horizontal components of motion, leading to two equations: Vy=Viy+ay*t and Vx=Vix+ax*t. The user initially calculated t as 0.87 seconds but faced confusion regarding the sign and setup of the problem, indicating a need for careful consideration of the direction of motion and the correct application of the equations.
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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