A question about acceleration and displacement

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SUMMARY

The discussion centers on calculating displacement for an object accelerating uniformly from rest. Given an acceleration of 7.09 m/s² north and a time of 7.50 seconds, the correct displacement can be calculated using the formula d = 1/2 * a * t². Substituting the values, the displacement is determined to be 197.625 meters north, correcting the initial miscalculation of 399 meters that did not account for acceleration.

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Hey everyone. I'm sort of new to the whole physics world (I'm doing it in highschol currently).
I have one question that I apparently can't figure out.

An object accelerates uniformly from rest. After 7.50s, the velocity is 53.2m/s north. What is the displacement?
I first did d=tv and got 399m north. But I know that isn't right because I didn't use any acceleration. I know that a=7.09m/s^2 north, but I'm not sure how to factor that into the 399m...
 
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try this equation:

d = v(0)*t + 1/2*a*t^2

v(0) is the initial speed which in this case is zero so the equation simplifies to:

d = 1/2*a*t^2

Plug in the acceleration you found (which is correct) and the time (7.5s) and you should get the correct answer.
 

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