The leader of a bicycle race is traveling with a constant velocity of 11.9 m/s and is 10.6 m ahead of the second-place cyclist. the second-place cyclist has a velocity of 9.80 m/s and an acceleration of 1.20 m/s2. how much time elapses before he catches the leader

6.3 sec

Explanation:

x = distance after which first-place bicyclist is caught

t = time taken to get caught

Consider the motion of the first-place bicyclist :

x = distance traveled by first-place bicyclist

t = time of travel

v = velocity = 11.9 m/s

distance traveled by first-place bicyclist is given as

x = v t

x = 11.9 t                                              eq-1

d = distance between first and second place bicyclist = 10.6 m

Consider the motion of the second-place bicyclist :

x' = distance traveled by first-place bicyclist = x + d = x + 10.6

t = time of travel

a = acceleration = 1.20 m/s²

v₀ = initial velocity = 9.80 m/s

Using the equation

x' = v₀ t + (0.5) a t²

x + 10.6 = 9.80 t + (0.5) (1.20) t²

using eq-1

11.9 t + 10.6 = 9.80 t + (0.5) (1.20) t²

t = 6.3 sec

they start with the same

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .