course phy 201 Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 10 cm/s to 13 cm/s as it travels 57.5 cm, then what is the average acceleration of the object?vo=10cm/s vf=13cm/s 'ds=57.5cm a=the change in velocity/the change in time but we also don't know time so to find time we will first need vAve so (vf+vo)/2 = (13+10)/2 =11.5cm/s this leads us to vAve='ds/'dt = 11.5cm/s = 57.5cm/'dt = (multiple both sides by 'dt to get into the numerater) 11.5cm/s * 'dt = 57.5cm (now divide both sides by 11.5 to isolate the 'dt) = 5 sec so we found vAve=11.5cm/s 'dt=5sec 'dv=3cm/s (vf-vo)=(13-10)(didnt work that one out up top sorry) a=3cm/s/5cm=.6cm/s Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 57.5 cm, starting from velocity 10 cm/s and accelerating at .6 cm/s/s. vf=vo+a'dt vf=10cm/s+(.6cm/s/s)(57.5cm) = 44.5cm/s 'ds=(vf+vo)/2*'dt 57.5cm=(44.5cm/s+10cm/s)/2*'dt = 57.5cm=27.25cm/s*'dt = (divide both sides by 27.25cm/s to isolate 'dt) = 2.11s 'ds is given at 57.5cm ; vo is given at 10cm/s; a is given at .6cm/s/s we have solved for vf and 'dt.