Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems: 41


The final distance between the two trains is $40m$.

Work Step by Step

Step-1: Let us calculate the acceleration of the first train from the graph. The slope of velocity-time graph is equal to acceleration. Therefore, the acceleration of the first train: $$a_1=\frac{rise}{run}=\frac{0-40}{5-0}=-8ms^{-2}$$ Step-2: From the relation, $$v^2=v_0^2+2ax$$ We substitute values to obtain: $$(0)^2=(40)^2+2\times (-8) \times x_1$$ $$\implies x_1=100m$$ So, the first train stops after covering $100m$, from the starting point. Step-3: Similarly, for the other train, acceleration is$$a_2=\frac{0-(-30)}{4-0}=7.5ms^{-2}$$ Here, $v_0=-30ms^{-1}$. This is because, each unit on the velocity-axis is $10ms^{-1}$. Here, we will use the same relation (as used in Step-2), $$(0)^2=(-30)^2+2\times 7.5\times x_2$$$$\implies x_2=-60m$$Here, the negative sign implies that the train moved towards direction of the other train. Step-4: Finally, the separation between the two trains is $200 - (100 + 60) = 40m$.
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