Question:
A particle starts from the origin at t = 0 with an initial velocity of 7.0 m/s along the positive x axis.?
?
2013-11-12 00:39:35 UTC
Please help, my teacher didn't really do a good job at explaining how to do these problems in class.

A particle starts from the origin at t = 0 with an initial velocity of 7.0 m/s along the positive x axis. If the acceleration is (-2.7 i + 3.7 j) m/s2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate.


velocity
_____i m/s + ______ j m/s

position
_______i m + _______ j m
Three answers:
Rahul
2013-11-12 00:57:41 UTC
the initial velocity can also be written as



U = 7 i + 0j



the particle is having negative acceleration in x direction with value 2.7 so the Vx will keep on changing with that. After time t the Vx will be



Vx = Ux + Ax * t = 7 - 2.7*t

The place where the particle will start reversing in x direction is when Vx becomes 0

Vx = 7 - 2.7*t = 0

t = 7/2.7 sec = 2.6 sec



Vy at the time when x is maximum



Vy = Uy + Ay t = 0 + 3.7 * 2.6 = 9.62 m/s



velocity = 0 i m/s + 9.62 j m/s



Using the formula

S = ut + at^2/2



X position when it is maximum



X = Ux t + Ax t^2/2 = 7 * 2.6 - 2.7 * 2.6^2/2 = 18.2 - 9.13 = 9.07 m



Y = Uy t + Ay t^2/2 = 0 + 3.7 * 2.6^2/2 = 12.51 m



position = 9.07 i m + 12.51 j m
?
2013-11-12 09:09:53 UTC
Integrate acceleration to find velocity function. integrate velocity function to fins position function,

i and j are unit vectors, not variables. "i" multiplies by "1" and points in the right/left directions, and "j" points in the up and down directions. the constants of integration are always the initial value of whatever function is obtained by integrating (ie in this case , integrating a(t) to get v(t) we add 7i because initial v is 7 m/s to the right.)



a(t) = -2.7 i + 3.7 j

v(t) = (-2.7 i + 3.7 j)t + 7i

x(t) = (1/2)(-2.7 i + 3.7 j)t^2 + (7i)t



to find the the time the max of x-coordinate of x(t) is reached, use its derivative v(t) to find a local maximum



v(t) = (-2.7 i + 3.7 j)t + 7i = 0; since were observing horizontal component we can ignore 3.7j to consider only x-coord velocity



t = (7/2.7)i

= 2.59 secs



x(t=2.59) = (1/2)(-2.7 i)(2.59)^2 + (7i)(2.59)

= (-9.056) + (18.13)

= 9.074 meters

y(t=2.59) = (1/2)(3.7 j)(2.59)^2

= 12.41 meters

we know x velocity is 0 at this point, or y velocity we can find with v(t), considering only j vector



v(t) = (3.7 j)t

=> v(2.59) = 3.7(2.59)j

= 9.58j m/s



velocity: 9.58j m/s

position: 9.07i + 12.41j







double check it im using the computer calculator i could have messef up
Priyankar
2013-11-12 12:08:46 UTC
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