question_answer

A cyclist covers a certain distance in 50 minutes riding his bicycle at a certain speed and in 40 minutes riding at a faster speed. If he rides his bicycle changing his speed from lower to higher speed every ten minutes, how long would he take to cover the distance. Consider both the cases that arise. Assume that the motion of the cyclists is uniform.

Answer:

Let the distance covered by the cyclists be ‘d’. Let \[{{V}_{1}}\]be the lower speed and \[{{V}_{2}}\]be the higher speed. \[\therefore {{v}_{1}}=\frac{d}{50}\]                                      ………(1) \[{{v}_{2}}=\frac{d}{40}\]                                             ………(2) Case 1:- Starts with lower speed \[({{v}_{1}}).\] Distance covered in first ten minutes with lower speed \[{{v}_{1}}=\frac{d}{50}\times 10=\frac{d}{5}\] Distance covered in next ten minutes with higher speed\[{{v}_{2}}=\frac{d}{40}\times 10=\frac{d}{4}\] Hence, in fourty minutes he covers \[=\frac{d}{5}+\frac{d}{4}+\frac{d}{5}+\frac{d}{4}=\frac{9d}{10}\] Distance remaining \[=d-\frac{9d}{10}=\frac{d}{10}\] He covers this distance with lower speed\[{{V}_{1}}\]in \[\frac{d/10}{d/50}=5\,\min \] \[\therefore \]Total time taken\[=40+5=45\,\,\min .\] For case 2, when he starts with higher speed we get 44 min.