Question

Two cars start in a race with velocities $u_1$$and$u_2$ and travel in a straight line with acceleration 'a' and b. If both reach the finish line at the same time, the range of the race is

• A. $\frac{2(u_1-u_2)}{(\beta -\alpha {)}^2}(u_1\beta -u_2\alpha )$
• B. $\frac{2(u_1-u_2)}{\beta +\alpha }(u_1\alpha -u_2\beta )$
• C. $\frac{2(u_1-u_2{)}^2}{(\beta -\alpha {)}^2}$
• D. $\frac{1u_1{u}_2}{\beta \alpha }$

Answer 1

The correct option is A $\frac{2(u_1-u_2)}{(\beta -\alpha {)}^2}(u_1\beta -u_2\alpha )$

$\begin{array}{c}S=ut+\frac{1}{2}\alpha t^2\\ \therefore X=u_{1}t+\frac{\alpha t^2}{2}\to \left(i\right)\end{array}$

For second car

$\begin{array}{c}X=u_{2}t+\frac{\beta t^2}{2}\to \left(ii\right)\\ now,\left(i\right)=\left(ii\right)\\ u_{1}t+\frac{\alpha t^2}{2}=u_{2}t+\frac{\beta t^2}{2}\\ t=\frac{2\left(u_1-u_2\right)}{\beta -\alpha }\\ Puttin\left(i\right)\\ X=\frac{2\left(u_1-u_2\right)\left(u_1\beta -\alpha u_2\right)}{{\left(\beta -\alpha \right)}^2}\end{array}$