Question 349044: Help me Please (Physics)
An object uniformly accelerates at a rate of 1.00m/s^2 east. While accelerating at this rate, the object is displaced 417.2m east in a time of 27.0s. What velocity did this object reach in this time?
& if you can, could you please help me to figure out when to use the right formula???
I also for the Formula d=vot+1/2 at^2 if you are supost to do .5 x a x t^2 divide d not sure how this one works (I am Confused with these formulas):(
Thank You and it would really really be most appreciated!!!
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! 29.0 m/s [east] is your answer
Velocity of an object with constant acceleration: vf = vi + at
In the above equation vf = final velocity of the object, vi = initial velocity of the object, a = acceleration, and t = time.
Displacement during constant acceleration: d = 1/2(vf + vi)t
The variables in the above equation are the same as in the previous equation.
Displacement when acceleration and time are known: d = vit + 1/2at2
The above equation is composed of two terms. The first term, vit, corresponds to the displacement of an object if it were moving with constant velocity, vi. The second term, 1/2at2, gives the displacement of an object starting from rest and moving with uniform acceleration. The sum of these two terms gives displacement of an object that starts with an initial velocity and accelerates constantly.
Displacement when velocity and acceleration are known: vf2 = vi2 + 2ad
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