I can't understand result in my textbook.

Question

I've the following equation in my physics text book: given that $$t=t_1+t_2+t_3,$$$$t_2=t_3$$ $$s_1 = v_1t_1,$$ $$s_2 =v_2t_2;$$ $$s_3 = v_3t_3,$$$$s_2+s_3=\frac{s}{2}=s_1$$ $$t= t_1 + 2t_2 \implies t_2=\frac{t-t_1}{2}.$$ From here starts the part that I don't understand: $$s_1=\frac{s}{2}=s_2+s_3= v_2t_2+v_3t_3=(v_2+v_3)\frac{t-t_1}{2}=v_1t_1$$ from here they jump to: $$t=\frac{(2v_1+v_2+v_3)t_1}{v_2+v_3}$$ I don't understand how they made this transition, I'm skipping some little detail and I can't find it.Any hits which will make it easier ?

Answer 1

Take the last two parts to the chain equality and divide both sides by $(v_2+v_3)/2$ to get $$t-t_1=\frac{2v_1t_1}{v_2+v_3}$$ then adding $t_1$ to both sides gives $$t=\left(\frac{2v_1}{v_2+v_3}+1\right)t_1$$

Now just simplify the term in parens by changing the $1$ to $\frac{v_2+v_3}{v_2+v_3}$

Answer 2

$$ (v_2+v_3)\frac{t-t_1}{2}=v_1t_1 \iff (v_2+v_3)(t-t_1)=2v_1t_1 $$ that, using distributivity, gives $$ (v_2+v_3)t-(v_2+v_3)t_1=2v_1t_1 \iff(v_2+v_3)t=2v_1t_1+(v_2+v_3)t_1 $$

can you do from this?

Answer 3

We have $$(v_2+v_3)\frac{t-t_1}{2}=v_1 t_1$$ This can be written as: $$(v_2+v_3)(t-t_1)=2v_1 t_1$$ Now, we can expand: $$v_2 t-v_2 t_1+v_3 t-v_3 t_1=2v_1t_1$$ And factorise: $$t(v_2+v_3)-t_1(v_2+v_3)=2v_1 t_1$$ Can you take it from here?

Answer 4

Starting with $$(v_2+v_3)\frac{t-t_1}{2}=v_1t_1\ ...$$

\begin{align} (v_2+v_3)\frac{t-t_1}{2}&=v_1t_1\\ t(v_2+v_3)-t_1(v_2+v_3)&=2v_1t_1 & \text{Multiply by 2 and distribute on LHS}\\ t(v_2+v_3)&=t_1(v_2+v_3)+2v_1t_1 & \text{Move $t_1(v_2+v_3)$ to RHS}\\ t(v_2+v_3)&=t_1(2v_1+v_2+v_3) & \text{Factor $t_1$ from RHS}\\ t&=\frac{t_1(2v_1+v_2+v_3)}{v_2+v_3} & \text{Divide by $v_2+v_3$} \end{align}