shiu planned a 210 km bike ride. However, he rode 5km per hour faster than he planned and finished his ride earlier than he planned. what is his average speed for his ride in km per hour?
So: S= 210/6 S= 35 km/h
I have no idea how you get 6, because there are 60 minutes in an hour but there is no information to say that it should be 6.
But yeah, that's how you do it.
Further, it can be interpreted in two ways. One is Shiu's speed is 5 km/h more than the planned and the other is Shiu accelerated 5 km/(h)^2 durring the ride.
Assuming planned speed S1
In first case, Shiu rode (S1 + 5). In this case, 210 = (S1+5) x (t1), where t1 is time taken
In the second case, Shiu's speed at a given time,t, is (5xt) + S1. In this case, assuming t2 is the time taken, ((S1 + 5t2 + S1) / 2) x t2 (i.e. area of trapezium under speed time graph) = 210.
In both of these cases, results are inconclusive due to lack of information.
Also, tried to see if it can be predicted from using hyperbolic functions i.e t1 = (210 / S1) and t2 = (210 / (S1 + 5)), i.e let S1 be variable as well (and form xy (i.e. t1 x S1) = C squard and use may be inverse function, domain and range for rearranging, i.e. S1 against t1, generally S against t in for both graphs) and tried to define some limits or using the area under defined limits as distance. I might have missed some thing because it has been very long time since I worked with hyberbolic functions, integration etc.
So, it may be worthwhile for some one who is hot trend on the hyberbolic functions, integration an functions etc. to try, and check if you can crack it assumong question is correct.