Abicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.0 m/s . two seconds later the bicyclist hops on his bike and accelerates at 2.0 m/s^2 until he catches his friend. a.) how much time does it take until he catches his friend (after his friend passes him)? b.) how far has he traveled in this time? c.) what is his speed when he catches up?

A.) t₁= 4.37 s

B.) d₁= 19.09 m

C.)

Explanation:

Bicyclist kinematics :

Bicyclist moves with uniformly accelerated movement:

d₁ = v₀₁*t₁ + (1/2)a₁*t₁²  equation (1)

d₁ : distance traveled by the cyclist

v₀₁: initial speed (m/s)

a₁: acceleration  (m/s²)

t₁:  time it takes the cyclist to catch his friend (s)

Friend  kinematics:

Friend moves with constant speed:

d ₂= v₂*t₂ Equation (2)

d₂ : distance traveled by the friend (m)

v₂: friend speed (m/s)

t₂ : time that has elapsed since the cyclist meets the friend until the cyclist catches him.

Data

v₀₁=0

a₁ = 2.0 m/s²

v₂ = 3.0 m/s

Problem development

A.) When the bicyclist catches his friend, d₁ = d₂=d  and t₂=t₁+2

in the equation (1) :

d = 0 + (1/2)2*t₁²  = t₁²

d = t₁²  equation (3)

in the equation (2) :

d = 3*(t₁+2) ₂ = 3*t₁+6

d =  3t₁+6 equation (4)

Equation (3) = Equation (4)

                t₁²  =  3t₁+6  

t₁² - 3t₁ - 6  = 0 ,we solve the quadratic equation

t₁= 4.37 s : time it takes the cyclist to catch his friend

B.) d₁ : distance traveled by the cyclist

In the Equation (2) d₁= (1/2)2* 4.37² = 19.09 m

C.)  speed of the cyclist when he catches his friend

force tends to be low and long diustance masses are high