Help with time/acceleration problem

[nathan scott]

Homework Statement

An object starts from rest at the origin and accelerates for a time t (x axis). Acceleration is constant. Object continues with same acceleration for an additional one second. The distance traveled during time t is one half the distance traveled during the one second interal' Find the time t.

Homework Equations

These are guesses. Total time = t+1.0s.
Total distance = x + 1/2x = 1.5x

The Attempt at a Solution

I haven't been able to figure this out. I keep going in circles.

 

[LowlyPion]

Welcome to PF.

What is the relationship between distance and time for an object under constant acceleration?

 

[nathan scott]

Welcome to PF.

What is the relationship between distance and time for an object under constant acceleration?

I've been playing with these equations: v = 1.5x/(t+1.0s)
a = (1.5x)/(t+1.os)²

Am I on the right path?

 

[LowlyPion]

I've been playing with these equations: v = 1.5x/(t+1.0s)
a = (1.5x)/(t+1.os)²

Am I on the right path?

I would prefer to see you rely on the relationship that x = 1/2*a*t²

 

[nathan scott]

I would prefer to see you rely on the relationship that x = 1/2*a*t²

I think I'm doing something fundamentally wrong because I try:

1.5x = 1/2* 1.5x/(t+1.0s)²*(t+1)²
But then I end up with 1.5x = 0.75x, which is obviously incorrect.
Something isn't clicking with me and this problem.

 

[LowlyPion]

Not quite.

You have 3/2*(1/2*a*t²) = 1/2*a*(t + 1)²

Just solve for t. Eliminating 1/2*a directly :

3/2 t² = (t + 1 )²

Take the square root of both sides ...

 

[nathan scott]

Not quite.

You have 3/2*(1/2*a*t²) = 1/2*a*(t + 1)²

Just solve for t. Eliminating 1/2*a directly :

3/2 t² = (t + 1 )²

Take the square root of both sides ...

I don't think it works. This ends up being : (the squared root of 6)/2*t = t + 1.
The x variable I was using in 1.5x was under the assumption that distance x + 1/2x = Total X. So I don't think I can plug 1/2*a*t² into the 1.5x, or 3/2x side of the equation. What do you think?

 

[LowlyPion]

The 1/2 a*t² is already the distance you call x.

The 3/2*x then is 3/2*(1/2*a*t²) and from the statement of the problem that's the same as !/2*a*(t + 1)²

As to the answer you got, I'd recheck your math.

 

[nathan scott]

I appreciate your help but I'm still confused. I'll keep pluggin away at it. I tried a different setup for time. I tried Total time = t, one portion of t (for distance x) would be 1.0s and the other distance (1/2*x)would be t-1.0s. I'm just not sure how to solve for t with the information supplied in the problem. I know the answer is 1.37s but I can't figure out how to arrive at that.

 

[LowlyPion]

I think I see the problem. I made the same mistake.

The statement of the problem says that

x_t = 1/2*(x_t+1 - x_t)

This means that

3x_t = x_t+1

not 3/2.