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.

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