If an object with uniform acceleration (or deceleration) has a speed of #12 m/s# at #t=0# and moves a total of 8 m by #t=1#, what was the object's rate of acceleration?

1 Answer
Nathan L.
Jul 2, 2017

#a_x = -8# #"m/s"^2#

Explanation:

We're asked to find te magnitude of the acceleration of an object with some given kinemtics quantities.

We know:

  • initial velocity #v_(0x)# is #12# #"m/s"#

  • change in position #Deltax# is #8# #"m"#

  • time #t# is #1# #"s"#

We can use the equation

#Deltax = v_(0x)t + 1/2a_xt^2#

Plugging in known values, and solving for the acceleration #a_x#, we have

#8color(white)(l)"m" = (12color(white)(l)"m/s")(1color(white)(l)"s") + 1/2(a_x)(1color(white)(l)"s")^2#

#8color(white)(l)"m" = (12color(white)(l)"m") + 1/2(a_x)(1color(white)(l)"s")^2#

#-4color(white)(l)"m" = 1/2(a_x)(1color(white)(l)"s")^2#

#-8color(white)(l)"m" = (a_x)(1color(white)(l)"s"^2)#

#a_x = color(blue)(-8# #color(blue)("m/s"^2#

The acceleration is in the negative direction, which means the object will keep slowing down in the positive direction until it eventually slows to a stop, and it then speeds up in the negative direction.

Related questions
See all questions in Acceleration
Impact of this question
1778 views around the world
You can reuse this answer
Creative Commons License