#$&* course Phy 121 Sept 26 11 pm Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 8 cm/s to 11 cm/s as it travels 38 cm, then what is the average acceleration of the object? v_0=8 cm/s v_f=11 cm/s 'ds=38 cm 'ds=(v_f-v_0)/2 *'dt 'dt=ds/(v_f-v_0)/2 =38cm/(8cm/s+11cm/s)/2 =4 sec 'dt=4 sec a= (v_f-v_0)/'dt (11cm/s-8cm/s)/4sec a=0.75 cm/s^2 Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 38 cm, starting from velocity 8 cm/s and accelerating at .75 cm/s/s. 'ds=28 cm v_0=8cm/s a=0.75cm/s^2 v_f^2=v_0^2+2a*'ds =8cm^2+2*.75cm/s^2*28cm v_f^2=121 cm/s v_f=11 cm/s 'dt=(v_f-v_0)/a =(11m/s-8m/s)/0.75cm/s^2 =4 sec "