Recent content by patflo

thank you very much!

may answer would be: for the equation of path i put in y=180t and x=10 (when y=80 x=10) then i get a quadratic function in t. the result for t= 0.4s i put that in v and get v=72i+180j for magnitude: v=193,7 m/s but as result there should be 201 m/s

ok i tried out this: d/dt (y-40)²=d/dt 160x => dx/dt=405t-90 v=vx*i+vy*j v=(405t-90)*i+180j but how can i find the magnitude? since in the equation for v there is still the variable "t".

Homework Statement When a rocket reaches an altitude of 40m it begins to travel along the parabolic path (y-40)^2=160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy=180m\s, determine the magnitudes of the rockets...

i will try this out..thanks!..i i will inform you if it worked!

yeah i know..but try to put s(1,5)=7m, s(8)=-36m so that would mean that i go 7m to the right, than 7m back (in total 14m) than -36m (so 50m total) and then to -30,5. total distance traveled approxiamtely= 55,5m but i have no analytiv solving for this problem...its only by trying out...

ok..now i got for position after 9s: s(9)=-30,5m that means the car traveled to the left if i suppose that the positive direction is to the right. Same thing with velocity v(9)= 10 m/s But what's about the total distance travelled? how can i solve that? if i put all the values from 0 to 9s...

My answer: v=integrate(a dt)=t²-9t+v0 t=0 therefore vo=10 v=t²-9t+10 s=inegrate(v dt) =t³/3-9t²/2+10t+s0 t=0 s=1 therefore s0=1 s=t³/3-9t²/2+10t+1 is that correct?

1. The acceleration of a particle along a straight line is defined by a= (2t-9)m/s² where t is in seconds. At t=0,s=1m and v=10m/s. When t=9s determine a) the particles position b) the total distance traveled and c) the velocity 2. I think: a=dv/dt; v= ds/dt 3. From a=dv/dt you...