Cyclist acceleration and distanced traveled problem

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SUMMARY

The discussion centers on a physics problem involving a cyclist who accelerates to reach a post office on time after realizing he is 30 seconds late. The cyclist initially travels at a constant speed of 1 m/s and accelerates at 0.3 m/s². The solution reveals that the time taken to reach the post office is 14.14 seconds, covering a distance of 44.14 meters. The equations used include kinematic formulas for displacement and velocity under constant acceleration.

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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 + volt + 1/2at^2
V=volt +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:

[tex]x_1 = V_0t = t[/tex]

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:

[tex]x_2=V_ot + 0.5at^2=t+0.15t^2[/tex]

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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