Asked 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!
Answered by Damon
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

Answered by bobpursley
vf^2=2ad solve for a.

b. vf=at solve for t

c. avg velocity= distance/time
Answered by Mason
You guys are the best ....thank you
Answered by Damon
You are welcome :)
There are no AI answers yet. The ability to request AI answers is coming soon!