Cyclist acceleration and distanced traveled problem

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A cyclist traveling at a constant speed of 1 m/s realizes he will be late to the post office by 30 seconds and accelerates at 0.3 m/s² to arrive on time. The equations of motion are used to determine the distance traveled and the time taken during acceleration. Without acceleration, the cyclist would cover 30 meters in 30 seconds, while with acceleration, the displacement is modeled as x = t + 0.15t². By equating the distances from both scenarios, the solution reveals that the time taken is approximately 14.14 seconds and the distance traveled is about 44.14 meters. This problem emphasizes the application of kinematic equations in solving real-world motion scenarios.
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Homework Statement


A cyclist moving at a constant speed of 1m/s realizes that he will be late to the post office by 30 seconds. Hence, he accelerates at a constant rate of 0.3m/second square so that he reaches just in time. Determine the time taken and the distance traveled.


Homework Equations


x=x0 + Vot + 1/2at^2
V=Vot +at
V^2=Vo^2 +2ax

The Attempt at a Solution


i have tried everything from equating this two equation still i get the wrong answer..
the answer is t=14.14s, x=44.14m...
please help hopefully you can show me the solution...
 
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Consider first the case where the cyclist DOES NOT accelerate. From his starting position (where he first realizes that he is 30 seconds behind schedule), the distance he has traveled after 't' seconds is:

x_1 = V_0t = t

Since his speed, 1 m/s, is known.

Now consider the second scenario where he DOES accelerate. His displacement after a given amount of time is given by:

x_2=V_ot + 0.5at^2=t+0.15t^2

Now you are told that if he does not accelerate, his is late by 30 seconds (i.e. he is 30m away from the post office). You can use this information to relate x1 and x2 and hence solve the equations to find the time and distance.
 
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