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mohdhm
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1)
A particle starts form the origin at t=0 with an initial velocity having an x compoent and a y component. the particle moves in the xy plane with an x component of acceleration only
a)
Determine the components of velocity vector at any time and the total velocity vector at any time
Given Info: Vxi=20m/s, Vyi=-15m/s, Ax=4.0m/s^2, Ay=0.
Solved.
Answer: Vf=Vxi+Vyi = [(20+4.0t)(i hat symbol?) - 15(j hat symbol?)]m/s
b) calculate the velocity and speed of the particle at t=5?
Ok the tricky part... both velocity and speed have the same magnitude because with 2 dimensions you have to use the pythagorean theorm to add the magnitude of the x and y components. I also concluded that vector = speed but with theta(direction).
solved the problem for part B.
c) determine the x and y coordinate of the particle at any time t and the position vector of this time.
xf=vit=1/2at^2= (20t+2.0t^2)m
yf=vit=(-15t)m
so the position vector at any time t is [(20t+2.0t^2)i -15tj]m
Then the example question says.. let's test it out. and gives me some values
at t=5, x=150, y=-75
what is the magnitude of the displacement?
rf=sqrt[(150^2+(-75)^2]=170m
"note that this isnot the distance that the particle travels in this time! Can you determine the distance from the available data?
Now how do you determine the distance at t=5 ? do you just add the vector components of distance? If someone can please answer that last question, i could rest my mind
A particle starts form the origin at t=0 with an initial velocity having an x compoent and a y component. the particle moves in the xy plane with an x component of acceleration only
a)
Determine the components of velocity vector at any time and the total velocity vector at any time
Given Info: Vxi=20m/s, Vyi=-15m/s, Ax=4.0m/s^2, Ay=0.
Solved.
Answer: Vf=Vxi+Vyi = [(20+4.0t)(i hat symbol?) - 15(j hat symbol?)]m/s
b) calculate the velocity and speed of the particle at t=5?
Ok the tricky part... both velocity and speed have the same magnitude because with 2 dimensions you have to use the pythagorean theorm to add the magnitude of the x and y components. I also concluded that vector = speed but with theta(direction).
solved the problem for part B.
c) determine the x and y coordinate of the particle at any time t and the position vector of this time.
xf=vit=1/2at^2= (20t+2.0t^2)m
yf=vit=(-15t)m
so the position vector at any time t is [(20t+2.0t^2)i -15tj]m
Then the example question says.. let's test it out. and gives me some values
at t=5, x=150, y=-75
what is the magnitude of the displacement?
rf=sqrt[(150^2+(-75)^2]=170m
"note that this isnot the distance that the particle travels in this time! Can you determine the distance from the available data?
Now how do you determine the distance at t=5 ? do you just add the vector components of distance? If someone can please answer that last question, i could rest my mind