blackhole123 Posted September 10, 2010 Share Posted September 10, 2010 I am using the equation for constant acceleration to find velocity. v^2=v(initial)^2+2a(x-x(initial)) Dropping a rock from 4 meters to the ground. Initial velocity is 0, a=9.8, and displacement is -4. So v^2=2(9.8)(-4), and it SHOULD equal=-8.85m/s (I know this is the correct answer, I already got it right on my homework). The velocity should be negative. But obviously my above answer is impossible because you can't take the square root of a negative. When using these constant acceleration equations I make the sign in the above equation negative if it is going upward, and positive if it is going downward. Do I apply the negative sign after I solve, to show the direction? That just doesn't seem right. Help! Link to comment Share on other sites More sharing options...

granpa Posted September 10, 2010 Share Posted September 10, 2010 dunno. consider the case where v_{initial}=0 x_{initial}=0 then v=sqrt(x) v is double valued. It can be positive or negative corresponding to t>0 and t<0 respectively Link to comment Share on other sites More sharing options...

blackhole123 Posted September 10, 2010 Author Share Posted September 10, 2010 Makes sense, and then I apply whether is is positive or negative based on whether it is increasing or decreasing? In these problems what should a be, 9.8 or -9.8? Does that change based on whether it is going up or down? Why don't I just post the whole problem: To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact? The answer is 1.26x10^3m/s^2 So I got the answer I already posted for descent, 8.85, then figured out the velocity for ascent, which is 6.26. Then the average acceleration is change in velocity over the time it was in contact with the floor. In order to get the right answer 8.85 had to be negative so you could have: (6.26+8.85)/.012=1.26x10^3. One of those velocities has to be negative (and it should be descent velocity), no matter how you cut it, and you have to take a square root to get both of them. I guess because velocity is the derivative of position, and if the graph of the function is decreasing then the graph of its derivative is negative. Link to comment Share on other sites More sharing options...

Janus Posted September 10, 2010 Share Posted September 10, 2010 I am using the equation for constant acceleration to find velocity. v^2=v(initial)^2+2a(x-x(initial)) Dropping a rock from 4 meters to the ground. Initial velocity is 0, a=9.8, and displacement is -4. So v^2=2(9.8)(-4), and it SHOULD equal=-8.85m/s (I know this is the correct answer, I already got it right on my homework). The velocity should be negative. But obviously my above answer is impossible because you can't take the square root of a negative. When using these constant acceleration equations I make the sign in the above equation negative if it is going upward, and positive if it is going downward. Do I apply the negative sign after I solve, to show the direction? That just doesn't seem right. Help! You seem to be mixing up signs. On one hand you are saying that the correct answer is negative, but then say that negative means going upward. You can't drop a something and have have it move upward, unless the acceleration itself is upward, and if upward is negative, then the acceleration is negative. You have to be consistent throughout the problem when applying signs. So assuming that the correct answer is negative, then that means that Up is positive and Down is negative. Therefore: v(initial) =0 x=0 x(initial) = 4 A = -9.8m/s² and v² = 0²+2(-9.8)(0-4) v² = 2(-9.8)(-4) = 78.4 v=sqrt(78.4) = (+/-)8.85 You have to decide on whether the answer is positive or negative based on the context of the problem. Link to comment Share on other sites More sharing options...

blackhole123 Posted September 10, 2010 Author Share Posted September 10, 2010 Thanks, I think I see my mistake. I was thinking of "a"'s sign as meaning whether it was accelerating or not, i.e. -9.8 meaning it is decelerating and positive meaning it is accelerating, hence even though I considered down as negative, I had a as +9.8 since it increases acceleration when it is going downward. Link to comment Share on other sites More sharing options...

swansont Posted September 10, 2010 Share Posted September 10, 2010 Thanks, I think I see my mistake. I was thinking of "a"'s sign as meaning whether it was accelerating or not, i.e. -9.8 meaning it is decelerating and positive meaning it is accelerating, hence even though I considered down as negative, I had a as +9.8 since it increases acceleration when it is going downward. The + or - sign is the vector information; that term in the equation is actually a dot product. You can choose down to be positive or negative if you wish, as Janus has indicated, but the acceleration and displacement will have the same sign in this problem, so the product will be positive. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now