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The position-time equation for a certain train is xf=2.8m+(8.1m/s)t+(2.9m/s^2)t^2. What is the initial velocity of this train? What is its acceleration?
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The position-time equation for a certain train is xf=2.8m+(8.1m/s)t+(2.9m/s^2)t^2.

What is the initial velocity of this train?
What is its acceleration?

User Tim McClure
by
7.6k points

1 Answer

7 votes
the formula without numbers is

xf = xi + vxi(t) + (1)/(2) (a) {t}^(2)
do the initial velocity would be
8.1 m/s
and the acceleration would be
2.9 m/s^2
User PgmFreek
by
8.4k points
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