Abicyclist 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)?

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.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//