On a highway, two buses A and B are running at the same velocity of magnitude 30 ms^-1. The brakes cause a deceleration or A ms^-2 in bus A and that of bus B is $\frac{30}{7} {ms}^{- 2}$ . In an emergency when driver of the front car applies brakes, immediately its rear light turns red and braking begins. In response, driver of the rear bus also applies brakes to avoid a collision with the front bus. Every driver takes 1 s to apply the brakes after he saw a need for it. If bus A ahead of bus B, then the minimum separation between the buses before driver of bus I applies the brake is x₁. If bus B is running ahead of bus A, then the minimum separation between the buses before the driver of bus B applies brake is x₂. The value of $\frac{x_{1}}{3 x_{2}}$ is _________.

Difficult

Solution

For x₁, If collision does not like place. The final velocities of both should be same.
$∴ v_{f} = 30 - \frac{30}{7} t = 30 - 3 (t - 1)$
$\begin{array}{l} \frac{30}{7} t = 3 t - 3 or \frac{30}{7} t - 3 t = - 3 \\ Or \frac{9 t}{7} = - 3 \Rightarrow t = - \frac{21}{9} \end{array}$
The negative of t indicates that both come in rest before collision. The distance moved by bus B before coming in rest is
$s_{1} = 30 \times 1 + \frac{30^{2}}{2 \times 3} = 30 + 150 = 180 m$
The distance moved by bus A before coming in rest is
$\begin{array}{l} s_{2} = \frac{30^{2}}{2 \times \frac{30}{7}} = \frac{30 \times 7}{2} = 105 m \\ ∴ x_{1} = s_{1} - s_{2} = 180 - 105 = 75 m \\ For x_{2}, v_{f} = 30 - \frac{30}{7} (t - 1) = 30 - 3 t \\ Or \frac{30}{7} t - \frac{30}{7} = 3 t or \frac{9}{7} t = \frac{30}{7} \\ ∴ t = \frac{10}{3} s \end{array}$
The positive of t indicates that before collision, velocities of both buses are non-zero but same.
So, the distance moved by A before collision is
$\begin{array}{l} s_{1} = 30 \times 1 + 30 \times (\frac{10}{3} - 1) - \frac{1}{2} \times \frac{30}{7} {(\frac{10}{3} - 1)}^{2} \\ s_{1} = \frac{265}{3} m \\ s_{2} = 30 \times \frac{10}{3} - \frac{1}{2} \times 3 \times {(\frac{10}{3})}^{2} = 100 - \frac{50}{3} = \frac{250}{3} \\ x_{2} = s_{1} - s_{2} = \frac{265}{3} - \frac{250}{3} = \frac{15}{3} = 5 m \\ Or \frac{x_{1}}{3 x_{2}} = \frac{75}{3 \times 5} = 5 \end{array}$

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