An object is traveling at a constant speed of 12.0 m/s when it slows to a stop. If it takes 10.0 m for the object to stop, what is the magnitude of its acceleration?
Using 2ad = Vf^2 - Vi^2
2(a)(10m)=(0m/s)^2-(12.0m/s)^2
20a=-144
a=-7.2 m/s^2
To find the magnitude of the object's acceleration, we need to use the equation that relates acceleration (a), initial velocity (v₀), final velocity (v), and displacement (d):
v² = v₀² + 2ad
In this case, the object starts with an initial velocity of 12.0 m/s and comes to a stop (v = 0). The displacement, or distance traveled during deceleration, is given as 10.0 m. We can rearrange the equation to solve for acceleration (a):
a = (v² - v₀²) / (2d)
Plugging in the known values:
a = (0 - 12.0²) / (2 * 10.0)
First, square the initial velocity:
a = (0 - 144.0) / (2 * 10.0)
Now, multiply 2 by 10.0:
a = -144.0 / 20.0
Finally, divide -144.0 by 20.0:
a = -7.2
Therefore, the magnitude of the object's acceleration is 7.2 m/s². Note that the negative sign indicates deceleration, or acceleration in the opposite direction of the initial velocity.