Question

In a relay race, runner a is carrying the baton and has a speed of 3.00 m/s. when he is 25.0 m behind the starting line, runner b starts from rest and accelerates at 0.100 m/s2. how long afterwards will a catch up with b to pass the baton to b?

Answer 1

The distance moved by runner a is calculated using the formula s = ut +$\frac{1}{2} at^2$

For runner a , u = 3 m/s, a = 0, s = 25 + x, where x is the distance traveled by runner b before they catch up.

So,  25 + x = 3*t

The distance moved by runner b is calculated using the formula s = ut +$\frac{1}{2} at^2$

For runner b , u = 0 m/s, a = 0.1$m/s^2$, s =  x

So, x = $\frac{1}{2} *0.1*t^2=0.1t^2$

Substituting in the previous equation, $25+0.05t^2=3t$

                         $t^2-60t+500 = 0\\ \\ (t-50)(t-10)=0\\ \\ t=50 seconds, t=10 seconds$

So they will catch up after 10 seconds.