Recent content by ere

i am faced with some homework in Resolve/C++ and encountered several problems ensures t1 = #t1 * #t2 -what do the pound signs before the variables mean? -what does it mean when one text "*" another text? hope someone can help. can't really find much resolve/C++ help online. thanks!

oops sorry i left that out. just edited the thread.

Homework Statement A sailboat is traveling east at 4.5ms-1. a suddent gust of wind gives the boat an acceleration a = 0.70 ms-2, 40 degress north to east. what are the boat's speed and direction 6s later when the gust subsides? Homework Equations V(@6s) = V(init.) + at The Attempt at a...

yep. thanks a lot!

i understand the part where there is only one accel. but how can i get the y(11s)? all i have is the position after a 4s freefall, and to start with the freefall has an initial velocity.

ok that was quite a mess. V(@11s) = V(init.) + At = 0 + (a - 9.8)11 = 11a - 107.8 therefore the distance the rocket fell after motor has stopped (ie ignoring distance already travelled) is: S = S(11s, ignore) + V(@11s)t + .5(-9.8)t^2 = 0 + (11a - 107.8)(4) + (-9.8)(8) = 44a - 431.2 - 78.4...

V(@11s)^2 = V(init.)^2 + 2AS(11s) S(11s) = sqrt((V(@11s)^2 - V(init.)^2) / 2A) = sqrt((V(@11s)^2 - 0) / 2(a - 9.8)) ^ that will be the distance traveled during the first 11 seconds, with gravitational effect, in terms of total acceleration S(15s) = S(11s) + V(@11s)t + .5At^2 = (the whole...

the gravity put an additional -9.8ms-2 to the acceration, so should i assume: 1. the acceration of the rocket's motor is actually a+9.8 in order to overcome the effect of the gravity? then the answer is 2.220+9.8 = 12.02 which is wrong according to the system. 2. when the motor stops, the only...

"launch a weather rocket" prbm Homework Statement A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 11 s, then the motor stops. The rocket altitude 15 s after launch is 4600 m. You can ignore any effects of air resistance. a) Acc. of...