Physics - Kinematics Question with Solution | TestHub

Let $u$ be the initial velocity. $\therefore v_1=u+a{t}_1,v_2=u+d\left(t_3+t_2\right)\mathrm{and}$ $v_3=u+a\left(t_1+t_3+t_3\right)$ Now, $v_1=\frac{u{w}_1}{2},\frac{u\left(u+a{t}_1\right)}{2}=u+\frac{1}{2}$ at, $v_1=\frac{v_1+c_2}{2}+u+a{t}_1+\frac{1}{2}a{t}_2$ $v_3=\frac{v_2+v_2}{2}=u+a{t}_1+a{t}_2+\frac{1}{2}a{t}_3$ So, $\left(y_1-v_2\right)=-\frac{1}{2}a\left(t_3+t_2\right)$ and $\left(v_2-v_3\right)=-\frac{1}{2}$ a $\left(t_2+t_3\right)$ $\therefore \frac{\left(v_1-v_2\right)}{\left(v_2-v_3\right)}=\frac{\left(t_1+t_2\right)}{\left(t_2+t_3\right)}$