course Phy201
Reason out the quantities v0, vf, v, vAve, a, s and t: If an object? velocity changes at a uniform rate from 11 cm/s to 15 cm/s as it travels 117 cm, then what is the average acceleration of the object? Manipulating the displacement equation algebraically you can obtain 117cm=13cm/s*?t. So now you divide 117 by 13 but you must invert it. So it will actually be 117cm*1s/13cm. This allows for the cm? to cancel out and you obtain 9s as the ?t. After finding the ?t you can figure out acceleration. A=(13cm/s)/9s=1.4444cm/s/s
Good reasoning on the average velocity and time interval.
Acceleration is, however, unrelated to average velocity. Dividing average velocity by time interval does not give you the acceleration.
What is the definition of acceleration, and how does it apply to this situation?
Using the equations which govern uniformly accelerated motion determine vf, v0, a, s and t for an object which accelerates through a distance of 117 cm, starting from velocity 11 cm/s and accelerating at .444 cm/s/s.
This problem will be quite long I will try to explain as best I can. You must first start off to find final velocity and this can be achieved by manipulating velocity algebraically to obtain Vf. Vf^2=v0^2+2*a*?t. So now this will become vf^2=11cm/s^2+2((.444cm/s/s)(117cm)^2).Now vf^2=224.896 2cm^2/s^2 take the sqrt224.896 2cm^2/s^2 and obtain Vf=14.997cm/s. Now that you have Vf you can obtain ?t. Manipulate the displacement formula to give you 117cm=(14.997cm/s+11cm/s)/2 * ?t. This will give ?t=9.005s.
Finally to obtain acceleration just take ?v/?t= .444cm/s/s
So now we have all the variables
Vf=14.997cm/s
V0=11cm/s
A=.444cm/s/s
?s=117cm
?t=9.00s
Good solution on the second problem.
Good work throughout except for your error on the acceleration in the first part.
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