A car traveling with a constant acceleration covers a distance of 60.9 m in 6.25

Question

A car traveling with a constant acceleration covers a distance of 60.9 m in 6.25 s. If its final speed is 15.7 m/s, what is its initial speed?

What is the car's acceleration?
Answer: 1.91e 00 m/s^2

20. [1 point, 10 tries]
At what distance from its starting point (x=0 m) was the car at rest?
Answer: 3.76e 00 m

Not sure how to do this. I tried using avg velocity = change in distance/change in time and got avg velocity of 9.744. And then I used avg velocity= vf v1/2 to get v(initial)=1.24 (which is wrong) and therefore acceleration which i got using Vf-V1/change in time =2.32 was wrong also.

HELPP

Explanation / Answer

15.7 = u + 1.9*6.25 u= 3.825 m/s 20.

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