Question

test your understanding
q4.1 according to the diagrams below, which relationships concerning the component and the resultant velocity are true?
at time t
at time t=0
a. $v_{0y}=v_{0}cos\theta_{0}$
b. $v_{x}=vcos\theta$
c. $\tan\theta=\frac{v_{x}}{v_{0}}$
d. $v_{0x}=v_{0}sin\theta_{0}$
e. $v_{y}=vsin\theta$
f. $v_{0}=sqrt{v_{0x}^{2}+v_{0y}^{2}}$
g. $v_{x}=sqrt{v^{2}+v_{0y}^{2}}$
h. $\tan\theta=\frac{v_{y}}{v_{x}}$
i. the angle, $\theta$, at some time t is always < $\theta_{0}$

Explanation:

Step1: Recall vector - component relationship

For a vector $\vec{v}$ with magnitude $v$ and direction $\theta$ with respect to the x - axis, the x - component $v_x = v\cos\theta$ and the y - component $v_y=v\sin\theta$. Also, $\tan\theta=\frac{v_y}{v_x}$. For the initial velocity vector $\vec{v}_0$ with magnitude $v_0$ and direction $\theta_0$ with respect to the x - axis, $v_{0x}=v_0\cos\theta_0$ and $v_{0y}=v_0\sin\theta_0$, and $v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$.

Step2: Analyze each option

• Option a: $v_{0y}=v_0\sin\theta_0$, not $v_0\cos\theta_0$, so a is false.
• Option b: By the vector - component formula, $v_x = v\cos\theta$, so b is true.
• Option c: $\tan\theta=\frac{v_y}{v_x}$, not $\frac{v_x}{v_0}$, so c is false.
• Option d: $v_{0x}=v_0\cos\theta_0$, not $v_0\sin\theta_0$, so d is false.
• Option e: By the vector - component formula, $v_y = v\sin\theta$, so e is true.
• Option f: By the Pythagorean theorem for vectors, $v_0=\sqrt{v_{0x}^2 + v_{0y}^2}$, so f is true.
• Option g: $v_x = v\cos\theta$, not $\sqrt{v^2 + v_{0y}^2}$, so g is false.
• Option h: $\tan\theta=\frac{v_y}{v_x}$, so h is true.
• Option i: There is not enough information to determine the relationship between $\theta$ and $\theta_0$ in general, so i is false.

Answer:

B. $v_x = v\cos\theta$
E. $v_y = v\sin\theta$
F. $v_0=\sqrt{v_{0x}^2 + v_{0y}^2}$
H. $\tan\theta=\frac{v_y}{v_x}$