# Help understanding why v is negative but moving positively

1. Feb 8, 2016

### Frankenstein19

1. The problem statement, all variables and given/known data

(This is a photocopy of the page, no real books were defiled in the process of studying :P)
2. Relevant equations

3. The attempt at a solution
In an example in my book, a problem reads that a car is moving in the positive x direction, the initial velocity is 15m/s and it takes 5 seconds to slow down to 5m/s. The example tells me the cars average acceleration is -2m/s^2. Then it says that in this case the acceleration is to the left, in the negative x direction evn though the velocity is pointing right. (I assume they mean instantaneous velocity) Now what I don't understand is how can the velocity be moving in the positive x direction but have a negative magnitude? If v1= 15 and vf=5, then the average v= -10. so if the signs tells us where the direction is, and this says it's negative, why is it moving in the positive x direction???

2. Feb 8, 2016

### SteamKing

Staff Emeritus
You're getting mixed up here. The car is said to be slowing down, which means that its velocity is decreasing over time. That's what the average acceleration of -2 m/s2 is telling you.

It does not mean the velocity is negative while the car is slowing, only that its magnitude is decreasing but still positive.

You're also not calculating the average velocity correctly. The formula for the average is $v_{avg} = \frac{v_i + v_f}{2}$

Last edited: Feb 9, 2016
3. Feb 9, 2016

### UchihaClan13

The above formula of average velocity being the arithmetic mean of the initial and final velocities works only if there's constant acceleration or retardation
Otherwise
I would use average velocity=Displacement/Time

4. Feb 9, 2016

### Ray Vickson

If you plot a graph of velocity $v$ vs. time $t$, then whether or not $v$ is positive or negative just depends on whether the graphed point $(t,v)$ lies above or below the $t$-axis.

The acceleration is the slope of the graph, so if acceleration is > 0 the slope is upward ($v$ increases as $t$ increases), while if acceleration is < 0 the slope is downward ($v$ decreases as $t$ increases). You seem to be confused between the value on the graph at $(t,v)$ and the slope of the graph at the point $(t,v)$.

5. Feb 10, 2016

### CWatters

It doesn't. It has positive Velocity (its always moving in the + ve x direction), but it has negative acceleration (=deceleration in this case). It slows from +15 to +5m/S.

Since when has the average of 15 and 5 been -10 ?

Perhaps you mean the change in Velocity? The change is negative which tells you the acceleration is negative but both the initial and final velocities are +ve.