ThinkPhysics

Problem: A ship moves along the equator to the east with velocity $v_0 = 30 \text{km/hour}$. The southeastern wind blows at an angle $\psi= 60^\circ$ to the equator with velocity $v = 15 \text{km/hour}$. Find the wind velocity $v'$ relative to the ship and the angle $\psi$ between the equator and the wind direction in the reference frame fixed to the ship.     Solution: We have to find $v-v_0$. In other words, we have to find $v+(-v_0)$. At this point, we can simply use vector addition, law of cosines, and law of sines to get that the velocity $$v'=39.7 \text{ at } 19^\circ \text{ from the east}.$$

Problem: Two particles, $1$ and $2$, move with constant velocities $v_1$ and $v_2$. At the initial moment their radius vectors are equal to $r_1$ and $r_2$. How must these four vectors be interrelated for the particles to collide?  Solution: Let the particles collide at point $A$ with position vector $r_3$.  Let $t$ is the time taken by the particles to reach $A$, and then we have $$r_3=r_1+v_1t=r_2+v_2t$$ $$\implies r_1-r_2=(v_1-v_2)t$$ $$\implies t=\frac{|r_1-r_2|}{|v_1-v_2|}.$$ From the last $2$ equations, we have $$r_1=r_2-(v_2-v_1)\frac{r_1-r_2}{|v_2-v_1|}$$, or $$\frac{r_1-r_2}{|r_1-r_2|}=\frac{v_2-v_1}{|v_2-v_1|},$$ which is the desired relationship. 

Problem: A car starts moving rectilinearly, first with acceleration $w = 5.0 \space \text{m}/\text{s}^2$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $w$, comes to a stop. The total time of motion equals $\tau = 25 \text{s}$. The average velocity during that time is equal to $\bar{v} = 72 \text{ km per hour}$. How long does the car move uniformly? Solution: Convert $72 \text{ km per hour}$ into $20 \space \text{m}/\text{s}$. Now, everything is in seconds and meters.  Since the rate of acceleration and deceleration are equal, the car must take the same amount of time and distance in this period.   Denote $t$ as the time in which the car moves uniformly. Then, the acceleration and deceleration each took $\frac{\tau-t}{2}$ time. So, we have $$\bar{v}\tau=2\left(\frac{1}{2}w\frac{(\tau-t)^2}{4}\right)+w\frac{(\tau-t)}{2}t$$ $$\implies t^2=\tau^2-\frac{4\bar{v}\tau}{w}$$ $$\implies t=\tau\sqrt{1-\frac{4\bar{v}}{w\tau}}=25\sqrt{1-\frac{4

Problem: A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_l$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.   Solution: Let $d$ be the total distance traveled by the point and let $t_1$ be the time taken to cover half of the distance. Then, let $2t$ be the time taken to cover the rest half of the distance.  Then, we have $$\frac{d}{2}=v_0t_1 \implies t_1=\frac{d}{2v_0}$$ and $$\frac{d}{2}=(v_1+v_2)t \implies 2t=\frac{d}{v_1+v_2}.$$ So, the mean velocity of the point is $$\frac{d}{t_1+2t}=\frac{d}{\frac{d}{2v_0}+\frac{d}{v_1+v_2}}=\boxed{\frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}}.$$