Required equations of motion in a plane
Position of a particle r(t) = r_o + v_0 t + \frac{1}{2}at^2
where, r_o \text{ and } r be the position vectors of the particle at time 0 and t and the velocities at these instants be v_0\ and\ v
Given that
Velocity of the particle v_0 = 5.0 \hat{i} m/s
Acceleration of the particle a = 3.0 \hat{i} + 2.0 \hat{j} m/s^2
a)
x- coordinate of the particle(x- component) x(t)= 84 m
y- coordinate of the particle = ?
At t = 0, r_0 = 0
Position of a particle r(t) = v_0 t + \frac{1}{2}at^2
r(t) = 5.0 \hat{i} t + \frac{1}{2} (3.0 \hat{i} + 2.0 \hat{j}) t^2
r(t) = 5.0 \hat{i} t + (1.5 \hat{i} + 1.0 \hat{j}) t^2
[math] r(t) = [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j} [/math]
Thus, position vector [math] r(t) = [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j}) [/math]
x- component x(t) = 5.0 t + 1.5 t^2 ..........................(1)
y- component y(t) = 1.0 t^2.............................(2)
84 = 5.0 t + 1.5 t^2 \because x(t)= 84
1.5t^2 + 5.0t - 84 = 0
0.5(t - 6)(3t + 26) = 0
Valid value for t = 6
Substituting the t value in equation (2)
y(t) = 1.0 t^2 = 1.0*6^2 = 36
Hence, y- coordinate of the particle = 36 m
b)
Speed of the particle =. change in position over time
v = \frac{dr}{dt}
[math] v = \frac{d}{dt}( [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j} )[/math]
[math] v= [5.0 + 3 t]\hat{i} + 2.0 t \hat{j} [/math]
Plugging in t = 6 in the above equation
Speed vector [math] v = [23]\hat{i} + 12\hat{j} [/math]
Magnitude of the speed v = \sqrt{v_x^2 +v_y^2} = \sqrt{23^2 + 12^2}
Therefore, speed of the particle v = 26m/s