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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

$r(t) = [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j}$

Thus, position vector $r(t) = [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j})$

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}

$v = \frac{d}{dt}( [5.0 t + 1.5 t^2]\hat{i} + 1.0 t^2 \hat{j} )$

$v= [5.0 + 3 t]\hat{i} + 2.0 t \hat{j}$

Plugging in t = 6 in the above equation

Speed vector $v = [23]\hat{i} + 12\hat{j}$

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