Easy AP Physics Q!!

Test Preparation
1

<p>A car traveling 12.5 m/s slows down at a constant acceleration of 0.50 m/s/s just by, "letting up on the gas". Calculate the distance the car coasts before coming to a stop.</p>

<p>The answer key (which by no means looks accurate, as its in a student's handwriting) claims:</p>

<p>V^2 = Vo^2 + 2ax</p>

<p>0^2 = (12.5)^2 + 2(-.50)X
X = 156.25m</p>

<p>But I don't think that's right.. don't I have to calculate the amount of time it takes to stop first using Vf = Vo+at (which i found to be 25 seconds) and plug it into this formula?</p>

<p>X = Xo + VoT + .5at^2</p>

<p>I got 468.75 meters with that formula..</p>

<p>And it seems kind of illogical that a coasting car going 45 km/h will come to a complete stop in 156 meters.</p>

2

<p>You should get the same answer of 156.25 with both methods. When you use
X = Xo + VoT + 0.5at^2
remember that 'a' has a negative sign, since you are decelerating.
X = (12.5)(25) - (0.5)(0.5)(25^2) = 156.25 meters .</p>

<p>(A coasting car would only face deceleration due to friction, which would be weaker than 0.5 m/s/s, I think...)</p>

3

<p>erm, thats so strange. i got 468.75 the first time around and ended up with 156.25 now. i guess i entered it into my calc wrong.</p>