A particle moves at a speed of 3.0m/s in the +x direction. Upon reaching the origin, the particle receives a continuous constant acceleration of 0.75m/s^2 in the -y direction. What is the position of the particle 4.0s later?
I also need the work arriving at the answer. Thanks so much!!
2006-09-19 09:32:39 · 3 answers · asked by spool 3 in Science & Mathematics ➔ Physics
Solve in 2 parts.
1. x- direction:
x=3.0*4=12.0 m (x=V*t when accelaration is zero)
2. y- direction
y=1/2*0.75*4^2=6 m (y=1/2*a*t^2 when initial velocity is zero)
3. (x,y)=(12,6)
2006-09-19 09:45:20 · answer #1 · answered by spenx01 1 · 0⤊ 1⤋
well, the x coordinate is obvious. X and Y direction motion are independant of each other, so the X position is just going to be t*v = X
the Y position requires a more heavy calculation. use the distance equation
d = (0.5) a * t * t = Y
since you're given velocity in the X direction and time, the first equation is nothing to solve.
the second equation is the same, you're given acceleration in Y direction and time, so you find out the distance in the Y direction and tada, you've now got both coordinates.
EDIT: HOLY CRAP SPENX, AT LEAST LET THE KID PLUG AND CHUG HER OWN HOMEWORK WITHOUT GIVING THE ENTIRE ANSWER.
2006-09-19 16:38:51 · answer #2 · answered by promethius9594 6 · 0⤊ 0⤋
lol why dont u do ur hw.....? hahah
2006-09-19 16:41:45 · answer #3 · answered by Help me 3 · 0⤊ 0⤋