A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he catches his friend. How much time does it take until he catches his friend (after his friend passes him)?

If the point of reference is located where the bicyclist is repairing his bicycle, his initial position is x0 = 0 when the persecution starts. The initial position of the friend, will be his position after 2 seconds of passing the point of reference:

x = x0 + vt = 0 m + 3.1 m/s * 2 s = 6.2 m

Since the bicyclist starts from rest, his initial speed is v0 = 0. Then:

x0 +v0 t + 1/2 a t² = x0 + v t

1/2 a t² = x0 +v t

1/2 a t² - v t = x0

1/2 a t² - v t - x0 = 0

1/2 * 2.4 m/s² t² - 3.1 m/s t - 6.2 m = 0

solving the quadratic equation:

t = 4 s

david8644 david8644 · Answer 2 · 2019-12-07T17:37:42+01:00

Answer:

The time it took them to meet is 3,9 seconds.

Explanation:

To solve this we have one incognita for both, which is the time that it took them to travel the same distance, the formula for the cycist that was traveling at constant speed is:

Distance=Speed*Time + 6,2 m

We add the 6.2 meters that the friend already covered in those two seconds.

And the formula for the cyclists that starts from rest would be:

Distance=1/2Acceleration*time^2

Since we want to find the point where the distance of both was the same we equalize the equations to eachother:

Speed*time + 6,2=1/2 Acc*time^2

3,1t+6,2 =1,2t^2

1,2t^2- 3.1t- 6,2=0

With this we have a quadratic equation that we solve with the general formula, see the image.

We get two results, remember that theres is no negative time so the answer would be the positive result.

Witht that we know that the time it took them to meet is 3,9 seconds.