# Homework Help: Solve this linear equation

1. Oct 9, 2009

### Osbourne_Cox

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.

2. Oct 9, 2009

3. Oct 9, 2009

### Osbourne_Cox

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

4. Oct 9, 2009

### calimechengr

Take a look at the second to last step. You have:

6.44*t = -5.6

5. Oct 9, 2009

### Osbourne_Cox

How do we have negative time though? (its the math to a physics problem)

6. Oct 9, 2009

7. Oct 9, 2009

### Osbourne_Cox

Degrees, and my calculator was set on degrees.

8. Oct 9, 2009

### 206PiruBlood

Maybe you should post the original physics problem. Is it possible you made an error setting the problem up ?

9. Oct 9, 2009

### Osbourne_Cox

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. Oct 9, 2009

### calimechengr

Sorry, I was not aware that 't' is time.

11. Oct 9, 2009

### Osbourne_Cox

Oh, it is very possible, ha ha. I am not good at this stuff at all.

12. Oct 10, 2009

### Staff: Mentor

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. Oct 10, 2009

### calimechengr

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.