i'm still wondering about the velocity for (a)...i feel like we were getting close and stopped working on that. can we work on that and then work on (c)?
for (a) is it v=20m/s??..that is what it comes out to be with the v=40m/s + (-10m/s^2)(2s)
and for (c) do I use the displacement...
just so there's no more confusion, the answer for b is the previous graph on the left...meaning the graph that curves IN when looking from the graph out b.c the graph on the left has a less steep slope than the one on the right.
am I correct?
you get 30m/s^2 right?
i took v=40m/s + (10m/s^2)(2s)=60m/s
"v=40m/s + (-10m/s^2)t and we want to find t when v=0. That is
0=40 + (-10m/s^2)t and solve for t. What do you get?"...i have no clue on the algebra on this!
ok so i think I'm solving for displacement then?...which would be the ave.velocity multiplied by time...and the ave velocity is 1/2(Vo + v)...so 1/2(40m/s + 60m/s)= 50m/s...if that is the ave. velocity then 50m/s(2s?) I'm confused on which time it wants...
i don';t understand how my vaules are wrong? the math is done right isn't it? when you throw the ball up the velocity is going down or slowing down until it reaches 0 at the turning point.
the value of A=10m/s^2...(this is what we are using other than 9.8m/s)
if the value is pos it is going...