Total distance q for mcat physics?

Question

a particle moving at 10m/s reverses its direction to move at 20m/s in the opposite direction. if this acceleration is -10m/s2, what is the total distance that it travels?

15m, 20m, 25m, or 30m....i got 15 m twice..but answer is 25...any help is appreciated.

Answer 1

I think your math is sound, I think you are just reading the problem wrong. The total displacement is 15m, but the question asks for distance traveled. To find this you must split the problem into two parts.

1. The distance it travels as the particle slows to 0m/s and then stops. and

2. The distance it travels as it picks up speed in the opposite direction.

For the first part you can see that it would take one second for the particle stop. Using the basic equation y1=y0+v0*t+(1/2)a*t^2, you get 5=(0)+(10)*(1)+(1/2)*(-10)*(1)^2, so it travels 5 meters in the positive direction before it stops.

Then to accelerate from 0m/s to -20m/s it takes another 2 seconds.

Plugging this in you get -20=(0)+(0)*(2)+(1/2)*(-10)*(2)^2, So it travels 20 meters in the negative direction as it accelerates.

5 meters forward and 20 meters back, makes for a total displacement of 15 meters, but a total distance of 25 meters. Hope this helps  

Answer 2

Slightly trick question because the particle changes direction and it sounds like they want to know the TOTAL distance traveled not just the NET distance.  Starting at 10 m/s and accelerating at -10 m/s^2 to -20 m/s means that it spends 1 second going forward (from 10 m/s to 0) and 2 seconds going in reverse from (0 to -20 m/s).  Calculate these separately:

1) forward: d = v0 t + 1/2 a t^2 = (10)(1) + (1/2)(-10)(1)^2 = 5m

2) reverse: d = v0 t + 1/2 a t^2 = (0)(2) + (1/2)(-10)(2)^2 = -20m

So you're right that the net distance traveled is 5m - 20m = -15m, but the total distance traveled is the sum of absolute values: 5m + 20m = 25m.

Like I said, kind of tricky.

First figure out the time: v_f = v_i + a t

-20 = 10 + (-10) * t -> t = 3 s

Then the distance:

x_f - x_i = v_0 t + 1/2 a t^2

= (10) * (3) + (1/2) (-10) * 3^2