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?

Answer 1

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