Recent content by bumblebee

Got it. Thanks.

I got -44 by multiplying -4.0 m/s * 11.0 s = - 44 m.

No problem. So for displacement would you use [v1 + v2 / 2] * t = d? [0 + -44]/2 x [11-1] = -22/10 = - 2.2 m? This does not seem right.

Okay let me try again. Is it 26 m? (26 + 16 = 42)?

I would not use it after the expiration date.

For area under the x-axis I got 34 so 34 + 16 = 50 m? This is what I did: 12 x (8-5) x (1/2) = 18 area of rectangle: 1 x 12 = 12 - [(1/2) * 1 * 8] = 8 4 x (11 - 9) = 8 18 + 8 + 8 = 34

How do I calculate final displacement and the total distance traveled in the graph above. I know displacement is independent of the pathway so the distance traveled on the last point is 44 m and the first point is 0 m. I also know for distance that you have to calculate the area of the...

Thanks! This helps!

Thanks, LowlyPion. Is it d=vt + 1/2at^2? I'm getting all these equations confused.

1. You are going toward home plate at speed of 5.9 m/s when you hit the dirt. You slide for 1.1 s just reaching the plate as you stop. a)What is the magnitude and direction of your acceleration? b) How far did you slide?I got 5.36 m/s^2 for the acceleration. Not sure if I did this right.