Recent content by aksarb

How do you get 200km/h to m/s? Do you just times that by 3.6?

Should it be positive because the 8.2m/s is going down due to the regualr sign conventions as left or down being negative? I conserded the sign conventions so the speed became -8.2m/s. Since there is the negative sign, it turned into a positive.

Ok thanks. Can you check the answer for this question? I did a=v2-v1/t = 3.1m/s + 8.2m/s / 3.0s = 3.8 m/s^2 t = v2-v1 / a = 3.1m/s + 8.2 m/s / 3.8m/s^2 = 3.0s

The question has the following data "After 3.0s, her velocity has changed to 3.1m/s." How come I didn't have to use the 3.0 seconds or the 3.1 m/s to get the answer?

Its ok, thanks for your help, I can finally relax :)

Ok, this question is asking for time right? So I just have to find acceleration from the information given then find time of the skier finally coming to a full stop after she originally fell?

Im not sure about the d. I do not have a displacement and I am not trying to find it, so how can I eliminate the d so the time can be by itself?

So, I would have to use basically get acceleration from v2-v1 then from there use t t = v2-v1 to get the answer? a

Thanks a lot this si a great help. Do you think you can help me with the other question I have posted?

So, it would be: D = (8.3m/s)(6.9s) + 1/2 (9.80/ms^2)(6.9s)^2 Work it out and the D would give me the height?

Homework Statement You throw a rock off a cliff, giving it a velocity of 8.3 m/s straight down. At the instant you released the rock, your hiking buddy started a stopwatch. The rock hit the water below exactly 6.9 seconds after you threw the rock. How high is the cliff above the water...

Homework Statement A skier is going 8.2m/s when she falls and starts sliding down the ski run. After 3.0s, her velocity has changed to 3.1m/s. How many seconds after she originally fell did she finally come to a stop? (Assume constant acceleration) Homework Equations Δd = 1(v1+v2)Δt...