A particle moves along the x axis according to the equation x = 2.03 + 2.90t ? 1

Question

A particle moves along the x axis according to the equation x = 2.03 + 2.90t ? 1.00t^2,where x is in meters and t is in seconds.
(a) Find the position of the particle at t = 2.55 s.

(b) Find its velocity at t = 2.55 s.

(c) Find its acceleration at t = 2.55 s.
Part 1 of 6 - Conceptualize

We are given an equation for x in meters as a function of t in seconds. Velocity is defined to be the
derivative of position x with respect to time t. Acceleration is defined to be the derivative of velocity with respect to time.

Part 2 of 6 - Categorize
To find the position of the particle, we evaluate the given expression at time t = 2.55 s. To find the velocity of the particle, we evaluate the derivative of the expression for x with respect to time t. To find the acceleration of the particle, we evaluate the derivative of the expression for velocity with respect to time t.

Part 3 of 6 - Analyze
(a) With the position given by x = 2.03 + 2.90t ? 1.00t^2,we substitute for t in the equation in order to find the position x at time t = 2.55 s.

x = 2.03 + 2.901_________??? ? 1.002 _________????
= 2.03 plus+_______??? minus? _______?????
= _______ m ????

Explanation / Answer

x= 2.03+ 2.90t - t^2 x= 2.03+ 7.395 - 6.5025 = 2.9225

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