Question

An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.85 cm. If its x-coordinate 2.45 s later is −5.00 cm, what is its acceleration?_________ cm/s2

Answer 1

vo = initial velocity = 11 cm/s

vf= final velocity = ?

a = acceleration = ?

t= time = 2.45 s

Δx = diplacement = xf -xi = -5 - 2.85 = -7.85 cm

Apply:

Δx = vot + 1/2 a t^2

-7.85 = 11 (2.45) + 1/2 (a) (2.45)^2

Solve for a:

-7.85 = 26.95 + 3 a

-7.85 - 26.95 = 3a

-34.8 / 3 = a

a = -11.6 cm/s^2