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Science question please help?
A ball rolls down an inclined plane with a constant acceleration of 2.5 m/s2:
a) how fast is the ball traveling after 3 seconds?
b) how far has the ball traveled in those 3 seconds?
c) how far has the ball traveled by the time its velocity is 15 m/s?
1 Answer
- GeronimoLv 78 years agoFavorite Answer
(a)
Vғ = (a • t) + Vi ... assuming the initial velocity = Vi = 0
Vғ = (2.5) • (3) + 0
Vғ = 7.5 m/sec ... final velocity after 3 seconds
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(b)
Dғ = (½) • a • t² + (Vi) • t + Di ... assuming the initial displacement = 0
... and the initial velocity = 0
Dғ = (½) • (2.5) • 3² + 0 + 0
Dғ = 11.25 meters ... final displacement after 3 seconds
Distance traveled = ∆D = Dғ – Di ... for straight line acceleration
∆D = 11.25 – 0
∆D = 11.25 meters ... distance traveled
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(c)
(Vғ)² – (Vi)² = 2 • a • (∆D)
(15)² – (0)² = 2 • (2.5) • (∆D)
∆D = 45 meters ... distance traveled