Posted by **Mason** on Monday, February 3, 2014 at 9:32pm.

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!

- Physics -
**Damon**, Monday, February 3, 2014 at 9:39pm
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

- Physics -
**bobpursley**, Monday, February 3, 2014 at 9:39pm
vf^2=2ad solve for a.

b. vf=at solve for t

c. avg velocity= distance/time

- Physics -
**Mason**, Monday, February 3, 2014 at 9:45pm
You guys are the best ....thank you

- Physics -
**Damon**, Monday, February 3, 2014 at 9:49pm
You are welcome :)

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