Question 2.SP.34: The motion of a point is described by the following equation......

Rewriting the given equations as

v_{x}=\frac{d x}{d t}=20 t+5 \qquad v_{y}=\frac{d y}{d t}=t^{2}-20

and integrating, the expressions for x and y are

x=10 t^{2}+5 t+C_{1} \qquad y=\frac{1}{3} t^{3}-20 t+C_{2}

To evaluate C_{1}, substitute x = 5 and t = 0 in the x equation. Then C_{1}=5. To evaluate C_{2}, substitute y = -15 and t = 0 in the y equation. Then C_{2}=-15. Substituting the values of C_{1} and C_{2}, the equations for displacement become

x=10 t^{2}+5 t+5 \qquad y=\frac{1}{2} t^{2}-20 t-15

Differentiate v_{x} and v_{y} to obtain the equations for acceleration:

a_{x}=\frac{d v_{x}}{d t}=20 \qquad a_{y}=\frac{d v_{y}}{d t}=2 t

Substituting t=2 \mathrm{~s} in the expressions for displacement, velocity, and acceleration, the following values are obtained:

x=\underline{55 \mathrm{~m}} \quad y=\underline{-53 \mathrm{~m}} \quad v_{x}=\underline{45 \mathrm{~m} / \mathrm{s}} \quad v_{y}=\underline{-16 \mathrm{~m} / \mathrm{s}} \quad a_{x}=\underline{20 \mathrm{~m} / \mathrm{s}^{2}} \quad a_{y}=\underline{4 \mathrm{~m} / \mathrm{s}^{2}}

The magnitudes and directions of the total displacement, velocity, and acceleration can be found by combining their components as before.