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?

a = (final v - initial v)/time

= (0 - 12)/t = -12/t so t = -12/a

d = Vi t +(1/2) a t^2

10 = 12 t + (1/2)a t^2

10 = 12(-12/a) + (1/2) a (144/a^2)

10 = -144/a + 72/a

10 a = -72

a = -7.2 m/s^2
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note
since velocity vs time is linear, you could use the average velocity during the stop
v = 12/2 = 6 m/s
so t = 10 / 6m/s = 5/3 second
a = -12 /(5/3) = - 36/5 = -7.2 m/s^2

To find the magnitude of acceleration, we can use the formula:

acceleration (a) = change in velocity (Δv) / time (t)

In this case, the object starts with a velocity of 12.0 m/s and slows down to a stop, so its change in velocity is given by:

Δv = 0 m/s - 12.0 m/s = -12.0 m/s

The time taken for the object to stop is given as 10.0 m. However, since the speed is constant during that time, we can use the formula:

Δv = a * t

Since the object's speed is decreasing, the acceleration is negative. Solving for acceleration, we have:

-12.0 m/s = a * 10.0 m

Dividing both sides of the equation by 10.0 m, we find:

a = -1.2 m/s²

Therefore, the magnitude of the object's acceleration is 1.2 m/s².