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**Example text**

When the bus starts, a car is at a distance b behind the bus and is moving in the same direction with constant speed U. Find the distance of the car behind the bus at time t after the bus has started for (i) 0 ~ t < VI/, and (ii) t ~ V/f Show that the car cannot overtake the bus during the period 0 ~ t < Vlf unless U 2 > 2bf Find the least distance between the car and the bu s in the case when U 2 < 2bfand U < V. State briefly what will happen if U 2 < 2bfand U> V. 18 The ends A and B of a beam of length 2a are in contact with a vertical wall and a horizontal floor respectively.

Also, the traces joining the horse and cart in Example 3, p. 31, were the means of exerting a force of magnitude T on the cart and an equal and opposite force on the horse. There are many similar examples where forces are tran smitted from one body to another through strings or rods . In this section we shall suppose that all the strings or rods considered are light, which means that their masses can be neglected. As shown in Fig. 6, consider a small element PQ of a light string or rod . Let the part of the string or rod to the left of P exert a force of magnitude S on PQ in the direction away from PQ , and let the part to the right of Q exert a force of magnitude Ton PQ in the direction away from PQ.

We now have to find rx and PN. Let the length of BP be x m. Since APB using Pythagoras' Theorem = The equation + G- x2 = T(cosrx - sin «) = W = AP = (4/5) m, Fig . 8 52 X)2 I = = x =i or t. cos z > sin « BP = (3/5) m. Sketch illustrating Example 10. Newton's Laws and Particle Motion = = 90°, AP> BP Let r m be the length of PN. Since sin ex = PN/AP = 5r/4, cos ex = PN/BP = 5r/3. sin ? ex + cos ? ex = I => r = (\2/25) cos ex - sin ex = (4/5) - (3/5) = (1/5) cos ex + sin ex = g = 9·81 ms ? => (7/5) length of PN is 0·48 m.