(4)

The slope of displacement time graph gives velocity

$\text{so, }{v}_1=\tan 30°; v_2=\tan 45°$

$\text{so, }\frac{v_1}{v_2}=\frac{\tan 30°}{\tan 45°}=\frac{1}{\sqrt{3}{\displaystyle \times }1}=1:\sqrt{3}$

(4)

Position of the car P at any time t, is

$x_P\left(t\right)=at+b{t}^2; v_P\left(t\right)=\frac{d{x}_P\left(t\right)}{dt}=a+2bt$                         ...(i)

Similarly, for car Q,

$x_Q\left(t\right)=ft-t^2; v_Q\left(t\right)=\frac{d{x}_Q\left(t\right)}{dt}=f-2t$                             ...(ii)

$\because v_P\left(t\right)=v_Q\left(t\right) \text{(Given)}$

$\therefore a+2bt=f-2t \text{or,} 2t(b+1)=f-a$

$\therefore t=\frac{f-a}{2(1+b)}$

(1)

Velocity of the particle is $v=At+B{t}^2$

$\frac{ds}{dt}=At+B{t}^2; \int ds=\int (At+B{t}^2) dt \therefore s=\frac{A{t}^2}{2}+B\frac{t^3}{3}+C$

$s(t=1 \text{s})=\frac{A}{2}+\frac{B}{3}+C; s(t=2 \text{s})=2A+\frac{8}{3}B+C$

$\text{Required distance}=s(t=2 \text{s})-s(t=1 \text{s})$

$=(2A+\frac{{\displaystyle 8}}{{\displaystyle 3}}B+C)-(\frac{{\displaystyle A}}{{\displaystyle 2}}+\frac{{\displaystyle B}}{{\displaystyle 3}}+C)=\frac{3}{2}A+\frac{7}{3}B$