Answer to Question #100471 in Mechanics | Relativity for Magdalene

Question

A particle with velocity v0= -2i+4j in meters per second at t=0 undergoes a constant acceleration a of magnitude a=3ms^-2 at an angle =130° from the positive direction of the x axis. What is the particles velocity v at t=5s, in unit vector notation and as a magnitude and an angle?

Answer 1

First of all we determine the unit vector with direction of acceleration. This may be done by expression $\hat n=\hat i\cdot cos\alpha+\hat j\cdot sin\alpha$ , where $\alpha=130\degree$. Vector of acceleration will be $\overrightarrow{a}=a\cdot \hat n$ . The equation of evolution of velocity is $\overrightarrow{V}=\overrightarrow{V_0}+\overrightarrow{a} \cdot t$ . In unit vector notation we can rewrite this equation in the form $\overrightarrow{V}=-2\hat i+4 \hat j+a\cdot t\cdot cos\alpha\cdot \hat i+a\cdot t\cdot sin\alpha\cdot \hat j$ . After reduction of similar terms we get $\overrightarrow{V}=(-2+a\cdot t\cdot cos\alpha )\cdot\hat i+(4+a\cdot t\cdot sin\alpha) \hat j$ . At $t=5s$ the velocity will be

$\overrightarrow{V}[m/s]=(-2+3\cdot 5\cdot cos (130\degree) )\cdot\hat i+(4+3\cdot 5\cdot sin(130 \degree)) \hat j=$ $-11.64\cdot \hat i+15.49\cdot \hat j$

It's magnitude is $V=\sqrt{(\overrightarrow{V}\cdot \overrightarrow{V})}=\sqrt{(-11.64)^2+(15.49)^2}=19.38\ m/s$ and an angle from the positive direction of the x axis is $\beta=acos(-11.64/19.38)=126.91\degree$ .

Answer: The partical velocity is $\overrightarrow{V}[m/s]=-11.64\cdot \hat i+15.49\cdot \hat j$ the magnitude $19.38\ m/s$ , and an angle from the positive direction of the x axis is $126.91\degree$ .

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