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.

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia