Question

two bicyclists begin a race at 10:00AM. They both finish the race 4 hours and 30 minutes later. prove that at some time during the race, the bicyclists are traveling at the same velocity

Answer 1

Two bicyclists begin a race at same time 10:00 AM .

The distance travelled by two bicyclists is same .

They both finish the race after 4 hours 30 min ( same time ) .

Let  s₁(t)  denotes the positon of first bicyclist at time t in minits .

Let  s₂(t)   denotes the positon of second bicyclist at time t in  minits .

Before starting the race thier positions is same ⇒ s₁(0) = s₂(0) 

4 hours 30 min = 270 min

Again after completion of race thier positions is same ⇒ s₁(270) = s₂(270) 

Let f (t) = s₁(t) -  s₂(t) 

Therefore f (0) = f (270) = 0              [ Since   s₁(0) = s₂(0)    and    s₁(270) = s₂(270) ]

The velocity of first cyclist is ds₁/dt ; the velocity of  second cyclist  is d s₂/dt.

Hence, 

df/dt = ds₁/dt - d s₂/dt 

According to , Rolle's  mean value Theorem  If you connect from f (a) to f (b) with a smooth curve, there will be at least one place where f ’(c) = 0 .

Therefore f(0) = f(270) = 0, Rolle's mean value Theorem , tells us that there is at least one value of t in (0, 270) for which df/dt = 0.

At this time, ds₁/dt - d s₂/dt  = 0, so at this time ds1/dt = ds2/dt.

Thus, there is at least one time in the race where the cyclists are traveling at the same velocity .