Given: - Speed of train A: \(v_A = 72 \, \text{km/h}\) - Speed of train B: \(v_B = 108 \, \text{km/h}\)
To convert the speeds from km/h to m/s:
\[ v_A = 72 \times \frac{1000}{3600} = 20 \, \text{m/s} \] \[ v_B = 108 \times \frac{1000}{3600} = 30 \, \text{m/s} \]
The relative velocity of train \(B\) with respect to train \(A\) is given by:
\[ v_{BA} = v_B - (-v_A) = v_B + v_A \]
Substituting the values:
\[ v_{BA} = 30 + 20 = 50 \, \text{m/s} \]
Since train \(B\) is moving towards the south and train \(A\) is moving towards the north, the relative velocity is considered negative:
\[ v_{BA} = -50 \, \text{m/s} \]
The velocity of the ground with respect to train \(B\) is simply the negative of the velocity of train \(B\) with respect to the ground:
\[ v_{\text{ground with respect to } B} = -v_B = -30 \, \text{m/s} \]
The velocity of train \(B\) with respect to \(A\) is \(-50 \, \text{m/s}\) and the velocity of the ground with respect to \(B\) is \(-30 \, \text{m/s}\).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).