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course Phy 121
Problem Number 3Determine the acceleration of an object which, after accelerating through a distance of 53 cm in 3.4 sec, is moving at 8.17647 cm/s.
known
ds = 53 cm
dt = 3.4 seconds
vf = 8.17647 cm/s
unknown
v0
a
The way I got the initial velocity was by taking ds/dt (53 cm / 3.4 s) and getting 15.6 cm/s..
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This is a good calculation, but it gives you the average velocity, not the initial velocity.
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I then got the acceleration by taking dv/dt ((8.2 cm/s - 15.6 cm/s) / 3.4 seconds) and got -2.2 cm/s^2.
v0 = 15.6 cm/s
a = -2.2 cm/s^2"
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Your procedure is good, except for the step where you calculated the average velocity and mistook it for the initial velocity.
That of course affects all your subsequent results.
The final velocity is about 8.2 cm/s and the average velocity is about 15.6 cm/s. What initial velocity, averaged with your final velocity, gives you the average velocity?
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