Question

A particle moves along the x axis according to the equation x = 1.91 2.99t-1.00e, where x is in meters and t is in seconds. (a) Find the position of the particle at t 2.90 s (b) Find its velocity at t- 2.90 s m/s (c) Find its acceleration at t 2.90s m/s?
My Note (a) Can the velocity of an object at an instant of time be greater in magnitude than the average velocity over a time interval containing the instant? Yes No (b) Can it be less? Yes O No Need Help?Resd
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Answer #1

A)

Position, X(2.90) = 1.91 + 2.99(2.9) - 1(2.9)^2 = 2.17 m

B)

Velocity, v(t) = 2.99 - 2t

V(2.9) = 2.99 - 2(2.9) = - 2.81 m/s

C)

Acceleration, a(t) = - 2 m/s^2

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