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

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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