wow..i was so in a hurry to reply on my previous post that i missed some operations. Now i got the "correct" answer. It's 3.1328 secs. I am correct in the assumption that the velocity after the acceleration phase will be the velocity to be used for the rest of the distance. Thanks for this post...
oopss..now i realized...the problem perhaps as i interpret it now is that whatever the velocity he will achieved after the acceleration phase, this will be his velocity for the rest of the distance. I tried this but still i got only t=2.39 secs not 3.1 sec. I am really getting very interested in...
oops.. i think i got myself confused... i was getting the g prallel to the inclined when the skier is still in horizontal. So for all these types of problems, we will treat all initial velocity and the g to be parallel to the inclined already? Thanks for showing me this. I am trying to refresh...
Hi, if the right answer is 3.1 sec. I tried to work this out backwards to check the "correct" answer.
Using the equation S=Vot +0.5at^2 to get the distance travelled with accleration for 3.1 secs,
s= (5.49)(3.1) + (0.5)(0.2)(3.1)^2 = 17.98 meters
And this would mean that the runner needs...
Thanks, however i find it the g parallel to the incline as g/sin11 = 51.36 /sec^2 but from the op the g is gsintheta = (9.8)(sin11) = 1.87 m/sec^2. I'm confused which is the correct g parallel to the incline that will add up to the initial speed/velocity of the skier.
hi , i thought this problem is to be solved by maxima/minima of calculus but following above suggestions i arrived at the following,
1) The final velocity when the car is accelerating is the initial velocity when it is decelerating (brakes applied).
2) So i used the formula suggested above...
hi, sorry for reviving this thread.
As for the computing for the length of the incline, it is correct to breakdown the g and 6.75 m/s initial velocity into components parallel to incline? Then use the formula vf^2=Vo^2 + 2aS to get the S?
hi ,
first, i try to get the t from the free fall equation d=.5gt^2, given d=1.2m and g=9.81 m/sec/sec. Once i get the t, i use this on the equation v = d/t, d = 0.41m and t from the first part.
Hello , sorry to interrupt but i am searching for kinematics problems and came across this thread and i am interested also in solving and i want to share what i have done.
The runner has only 180 seconds left to complete the run with a distance of 1100 m. I used two formula here, 1 with...
Hi Kurdt ,
I am trying to search the forum for average velocity formula. In this problem example. What if the person in this example runs in straight line (north) instead of south. What is the correct computation for the average velocity? Is it (6.5 km/hr + 18.5 km/hr)/2 = 12.5 km/hr North or...