Asst6 rand

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course phy 201

4/5 250pm

Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 9 cm/s to 12 cm/s as it travels 84 cm, then what is the average acceleration of the object?Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 84 cm, starting from velocity 9 cm/s and accelerating at .375 cm/s/s.

dv=vf-v0

12cm/s-9cm/s=3cm/s

vAve=(v0+v0)/2

(9cm/s+12cm/s)/2=10.5cm/s

dt=ds/vAve

84cm/10.5cm/s=8s

aAve=dv/dt

3cm/s/8s=.375cm/s^2

Equations:

vf^2=v0^2+2a*dt=

vf^2=9^2+2(.365)84=

vf^2=81+63

vf=sqrt144

vf=12

vf=v0+a*dt

12=9+.375dt

3=.375dt

8=dt

&#Your work looks very good. Let me know if you have any questions. &#