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
Jul 2, 2017

Answer:

#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.