at t=0, a particle moving in the xy plane with constant acceleration has a velocity of vi=(3.00i-2.00j) m/s and is at the origin. At t=3.00s, the particles velocity is v=(9.00i+7.00j). Find (a) the acceleration of the particle and (b) its coordinates at any time t.

a = ∆v/∆t = ((9i+7j)-(3i-2j)/3 = 2i+3j

v = (2i+3j)t + 3i-2j = (2t+3)i + (3t-2)j
s = 1/2 (2i+3j) t^2 + (3i-2j) t + C
since s(0) = 0, C=0. and
s = 1/2 (2i+3j) t^2 + (3i-2j) t = (t^2+3t)i + (3/2 t^2 - 2t)j

Thanks

Its coordinats at any times t

Question

To find the acceleration of the particle, we can use the formula for acceleration:

a = (vf - vi) / t

Where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken.

Given:
vi = (3.00i - 2.00j) m/s
vf = (9.00i + 7.00j) m/s
t = 3.00 s

Substituting the values:

a = ((9.00i + 7.00j) - (3.00i - 2.00j)) / 3.00

Simplifying further:

a = (6.00i + 9.00j) / 3.00

a = 2.00i + 3.00j

Therefore, the acceleration of the particle is a = (2.00i + 3.00j) m/s².

To find the coordinates of the particle at any time t, we can use the equations of motion:

x = xi + vix * t + (1/2) * ax * t^2
y = yi + viy * t + (1/2) * ay * t^2

Where x and y are the coordinates, xi and yi are the initial coordinates (which are both zero in this case), vix and viy are the initial velocities in the x and y directions, respectively, ax and ay are the acceleration components in the x and y directions, respectively, and t is the time.

Given:
xi = 0
yi = 0
vix = 3.00 m/s
viy = -2.00 m/s
ax = 2.00 m/s²
ay = 3.00 m/s²

Substituting the values:

x = 0 + 3.00 * t + (1/2) * 2.00 * t^2
y = 0 + (-2.00) * t + (1/2) * 3.00 * t^2

Simplifying further:

x = 3.00t + t^2
y = -2.00t + (3/2) * t^2

Therefore, the coordinates of the particle at any time t are x = 3.00t + t^2 and y = -2.00t + (3/2) * t^2.