Question 2.SP.31: The x and y components of the displacement in meters of a po......
The x and y components of the displacement in meters of a point are given by the equations
x = 4t² – 3t \qquad y = t³ – 10
Determine the velocity and acceleration of the point when t = 2 s.
The velocity components obtained by differentiation are
v_{x}=\frac{d x}{d t}=8 t-3 \qquad v_{y}=3 t^{2}
At t=2 \mathrm{~s}, v_{x}=13 \mathrm{~m} / \mathrm{s} and v_{y}=12 \mathrm{~m} / \mathrm{s}. Hence,
v=\sqrt{\left(v_{x}\right)^{2}+\left(v_{y}\right)^{2}}=\underline{17.7 \mathrm{~m} / \mathrm{s}} \qquad \text { and } \qquad \theta_{x}=\tan ^{-1} \frac{v_{y}}{v_{x}}=\tan ^{-1} \frac{12}{13}=\underline{42.7^{\circ}}
where \theta_{x} is the angle between the total velocity and the x axis.
A second differentiation yields the acceleration components:
a_{x}=\frac{d v_{x}}{d t}=8 \qquad a_{y}=\frac{d v_{y}}{d t}=6 t
At t=2 \mathrm{~s}, a_{x}=8 \mathrm{~m} / \mathrm{s}^{2} and a_{y}=12 \mathrm{~m} / \mathrm{s}^{2}. Hence,
a=\sqrt{\left(a_{x}\right)^{2}+\left(a_{y}\right)^{2}}=\underline{14.4 \mathrm{~m} / \mathrm{s}^{2}} \qquad \text { and } \qquad \phi_{x}=\tan ^{-1} \frac{a_{y}}{a_{x}}=\tan ^{-1} \frac{12}{8}=\underline{56.3^{\circ}}