help me with this?

theory of relative velocity….The cyclist would have travelled a distance of 16*5/4 = 20 kms..

relative velocity = 56-16 = 40 km/hr

Time taken to catch up = 20/40 = 0.5 hrs

Therefore, the motorist will catch up with the cyclist at 10:45, and at a distance of 16*0.5+20 = 28 kms from A

RATE MY ANSWER

That means that at 10:15 (after 1.25 hours have elapsed), the cyclist will have already travelled 16 + (1/4) 16 = 16 + 4 = 20 km. … 16 km after 1 hour has elapsed and another 4 miles after another 15 minutes passed… (since 15 minutes is 1/4 of an hour… then a quarter of 16km is 4 miles)

let x = time that cyclist traveled

Distance of cyclist = distance of motorist

(16km/h)(1.25 h + x) = (56 km/h)(x)

because Rate C times cyclist time = Distance Cyclist

and Rate M time motorist time = Distance Motorist….

We also know that the cyclist will have been on the road 1.25 hours longer than the motorist…

(16km/h)(1.25 h + x) = (56 km/h)(x) … now solve for x…

16(5/4) + 16x = 56x

20 = 40x

x = 1/2

x = 1/2 hr = 0.5h…

Which means that at 10:45 ( 10:15 plus 0.5 hour) and at a distance of 28 km… because (56km/h)(0.5 h) = 28 km…, the motorist and cyclist will be side-by-side…