What Is the Particle's Speed When Its X Coordinate Reaches 15m?

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Anthonyphy2013
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Homework Statement


At t=0, a particle leaves the origin with the velocity of 9.0m/s in the positive y direction and moves in the xy plane with a constant acceleration of ( 2.0i-4.0j) m/s ^2 . At the instant the x coordinate of the particle is 15 m , what is the speed of the particle ?


Homework Equations



V=u+aT , X= ut+ .5at

The Attempt at a Solution


X coordinate: X=ut+.5at
15= 0+ 2t
find t and plug it in the y - coorodinate
y coordinate : V=9-4t
Is that corrected ?
 

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Anthonyphy2013 said:

Homework Statement


At t=0, a particle leaves the origin with the velocity of 9.0m/s in the positive y direction and moves in the xy plane with a constant acceleration of ( 2.0i-4.0j) m/s ^2 . At the instant the x coordinate of the particle is 15 m , what is the speed of the particle ?


Homework Equations



V=u+aT , X= ut+ .5at
This is incorrect. X= ut+ .5at^2

The Attempt at a Solution


X coordinate: X=ut+.5at
15= 0+ 2t
find t and plug it in the y - coorodinate
y coordinate : V=9-4t
Is that corrected ?
You seem to be completely misunderstanding the notation. V is the vector velocity. It is NOT the y coordinate. X is the vector position, it has both x and y components. It is NOT the x coordinate.

If the initial speed is u (which itself a vector with x and y components) then V= u+ at. Here, you are told that "At t=0, a particle leaves the origin with the velocity of 9.0m/s in the positive y direction and moves in the xy plane with a constant acceleration of ( 2.0i-4.0j) m/s ^2" so that u= <0, 9> and a= <2, -4>. V= <0, 9>+ <2, -4>t= <2t, 9- 4t>. Then X= <t^2, 9t- 2t^2>.

The x coordinate will be 15 when t^2= 15 or t= sqrt(15). Then y= 9t- 2t^2= 9sqrt(15)- 30.
 
HallsofIvy said:
This is incorrect. X= ut+ .5at^2


You seem to be completely misunderstanding the notation. V is the vector velocity. It is NOT the y coordinate. X is the vector position, it has both x and y components. It is NOT the x coordinate.

If the initial speed is u (which itself a vector with x and y components) then V= u+ at. Here, you are told that "At t=0, a particle leaves the origin with the velocity of 9.0m/s in the positive y direction and moves in the xy plane with a constant acceleration of ( 2.0i-4.0j) m/s ^2" so that u= <0, 9> and a= <2, -4>. V= <0, 9>+ <2, -4>t= <2t, 9- 4t>. Then X= <t^2, 9t- 2t^2>.

The x coordinate will be 15 when t^2= 15 or t= sqrt(15). Then y= 9t- 2t^2= 9sqrt(15)- 30.

That means I just consider the x and y component of the acceleration and put them to find the velocity of y component