Relative velocity and accel. question? Can't figure out what velocity

The question in question is number 2.

The velocity must be 2, right? How would I go about getting a different number?

The time for the phone to get to the edge is 10 seconds, so that means they have to get there in 10 seconds, too. If they were to start immediately after they miss then they only need to be going the same velocity, right?

Where am I going wrong and how to get the right number?

Thank you for any replies

EdIt: i got it lol i used d=0.5at^2 my bad thanks yall lolol
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Aug 2010
Problem 2 does not ask anything about velocity, it asks for acceleration. The phone passes you at 2 m/s. After "t" seconds, it will have gone 2t meters. If you start after it with constant acceleration, a, in "t" seconds, you will have gone $(a/2)t^2$ meters. You will catch up to the phone when those two distances are the same, $(a/2)t^2= 2t$. That quadratic equation has two solutions. One of them is t= 0 when the phone first passed you. The other solution is when you catch up to it. That will, of course, depend on a. Then you need to put that value of t into 2t to find the distance the phone goes before you catch it. That will, again, depend on a. What must a be so that the distance is less than 20 m?
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