Answer:
A) t = 7.0 s
B) x = 25 m
C) v = 10 m/s
Explanation:
The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
A)When both friends meet, their position is the same:
x bicyclist = x friend
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:
Position of the friend after 2 s:
x = v · t
x = 3.6 m/s · 2 s = 7.2 m
Then:
1/2 · a · t² = x0 + v · t v0 of the bicyclist is 0 because he starts from rest.
1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t
1 m/s² · t² - 3.6 m/s · t - 7.2 m = 0
Solving the quadratic equation:
t = 5.0 s
It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.
B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.
x = v · t
x = 3.6 m/s · 7.0 s = 25 m
(we would have obtained the same result if we would have used the equation for the position of the bicyclist)
C) Using the equation of velocity:
v = a · t
v = 2.0 m/s² · 5.0 s = 10 m/s
