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