course PHY 201
Determine the acceleration of an object whose velocity is initially 18 cm/s and which accelerates uniformly through a distance of 55 cm in 4.2 seconds.Vi = 18 cm/s
`ds = 55 cm
`dt = 4.2s
[(vf+vi)/2] * `dt = `ds
[(vf+vi)/2] = `ds/`dt
Vf+vi = 2(`ds/`dt)
Vf = 2(`ds/`dt) –vi
Vf = 2 (55/4.2) – 18
Vf = 8.2 cm/s
(vf – vi) / 4.2 = a
A = 8.2 – 18/4.2
A = -2.3 cm/s^2
Very good solution.
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