Question

In a relay race, runner A is carrying the baton and has a speed of 2.60 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0600 m/s2. How long afterwards will A catch up with B to pass the baton to B?

Answer 1

Let the velocity of A is v = 2.60 m/s

Let the acceleration of B is a = 0. m/s²

Let the time taken by A to catch B be t

This means both A and B move for same time

A has to move 25 m to reach the starting point of B

After this mark both A and B move same distance

Let this distance be S and this means that A has travelled 25 m more than B to catch it

Distance moved by B is S = u*t + ½*a*t²

But u = 0 as B is initially at rest

So S = ½*a*t²

Let the total distance crossed by A be S_a such that

   S_a = v *t

We know that A has moved 25 m more than B so we can write

S + 25 = S_a

½ * a * t² + 25 = v * t ………………. (1)

½*0.06 *t² + 25 = 2.60* t

0.06t² - 5.20t + 50 = 0

On solving we get t = 11 and 75.65

we need the least time for A to reach B , so t must be 11 second

So A gives the baton to B in t = 10 sec

We can verify by substituting t in ( 1 )

½ * 0.06*(11)² + 25 = 2.60*11

28.6 = 28.6

So we can conclude that A gives baton to B in t = 11 s