#$&* course Phy 121 Determine the acceleration of an object whose velocity is initially 18 cm/s and which accelerates uniformly through a distance of 55 cm in 4.2 seconds. v_0=18cm/s 'ds=55 cm 'dt=4.2 s 'ds=v_0*'dt+(1/2)a*'dt^2 55 cm=(18cm/s)*(4.2 s) + 1/2a*(4.2 s)^2 55cm=75.6cm+.5a(17.64s^2) -19.6cm=.5a(17.64s^2) divide both sides by 17.64 s^2 -1.11cm/s^2=.5a acceleration=-2.22cm/s^2