Reading acceleration vs. time graphs

[exi]

Homework Statement

http://img265.imageshack.us/img265/4572/graphrj8.jpg

(The particle starts from rest at the origin.)

1: What is the velocity at t = 5s?

2: What is the position at t = 5s?

The Attempt at a Solution

I'm thinking -2 m/s velocity and -8 m position at t = 5s, but would like someone to double-check my understanding of reading velocity and position from this graph. The negative acceleration at origin combined with that instantaneous change at t = 1 have me a bit unsure of myself.

 

[Astronuc]

The zero acceleration simply means that the speed is constant after 1 sec, i.e. there is no further acceleration.

So the maximum speed (magnitude of velocity) is achieved at 1 sec.

Starting from rest with a negative acceleration, this would imply acceleration in the negative direction. Normally negative acceleration would imply a deceleration in the + direction.

 

[exi]

The zero acceleration simply means that the speed is constant after 1 sec, i.e. there is no further acceleration.

So the maximum speed (magnitude of velocity) is achieved at 1 sec.

Starting from rest with a negative acceleration, this would imply acceleration in the negative direction. Normally negative acceleration would imply a deceleration in the + direction.

yes sir, so does my thinking at t = 5 sound correct, or am I a bit off?

 

[Astronuc]

What is the velocity at 1 sec? It is constant thereafter.

 

[exi]

Either -1 or -2 m/s. That change at t=1 is what's making me a little unsure as to whether there's an extra increment of acceleration.

So velocity at t=5 is either -1 or -2, and position is... I'm still guessing -8 based on what I mentioned in the original post, but that depends on what I'm asking here, as well.

 

[Astronuc]

Well, for 1 s, the acceleration is constant at -1 m/s².

v = a*t = -1 m/s² * 1 s = ?

At t = 1 s, the acceleration drops to zero. That is a discontinuity. Since a = 0, v is constant and equal to v(t= 1s).

 

[exi]

Okay, so at t = 5, velocity is -1 m/s.

Oh wow, I was way off on that. Am I correct in using:

$$\Delta x = V_ot + \frac{at^2}{2}$$

twice, once for t(0,1) and t(1,5), adding the results, and getting -4.5m for position @ t=5s?

 

[Astronuc]

The problem can be broken into two time intervals, yes. 0 to 1s, with constant acceleration, the 1 to 5s, at constant velocity.

For the first part of the problem, v_o = 0, a = -1 m/s², over the time interval 0 to 1s.

For the second part of the problem, v_o = - 1 m/s, and a = 0, for t = 1 s to 5 s.

and -4.5m for position @ t=5s is correct.