Physics
posted by Mason .
An object at rest accelerates through a distance of 1.5 m at which point the instantaneous velocity of the object is 3.5m/s. Determine:
a. The average acceleration of the object
b. The time it took the object to travel 1.5m
c. The average velocity of the object during this period of motion
I saw this question was posted below, but not answered yet. I was also curious on how to do it. Thanks for the help!

a = change in velocity /change in time
= 3.5/t
what is t?
average speed is (3.5 + 0)/2 = 1.75 (that is part c)
1.75 t = 1.5 meters
t = .857 second (that is part b)
so
a = 3.5 / .857 = 4.08 m/s^2 ( part a)
when acceleration is constant, average speed during acceleration is (initial speed + end speed) /2
which saves a lot of computation 
vf^2=2ad solve for a.
b. vf=at solve for t
c. avg velocity= distance/time 
You guys are the best ....thank you

You are welcome :)
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