Li L.

asked • 10/07/19# PHYSICS HELP - A particle moves along the x axis..

A particle moves along the *x* axis according to the equation *x* = 2.08 + 3.01*t* − 1.00*t*^{2},

where *x* is in meters and *t* is in seconds.

(a) Find the position of the particle at *t* = 3.40 s.

(b) Find its velocity at *t* = 3.40 s.

(c) Find its acceleration at *t* = 3.40 s.

## 2 Answers By Expert Tutors

Have you taken calculus?

If yes:

a) plug in t=3.4, x is your position x=2.08 + 3.01(3.4) − 1.00(3.4)^{2}

b) take the derivative of x because the rate of change of position is velocity

x=2.08 + 3.01*t* − 1.00*t*^{2}

v=3.01-2t plug in t=3.4s v=3.01-2(3.4)

c) take the derivative of v because the rate of change of velocity is acceleration

v=3.01-2t

a=-2

acceleration is constant so it is always -2

if you havent taken Calculus use these two kinematics expressions

x=x_{0} + v_{0}*t* + 0.5a*t*^{2}

v=v_{0}+at

Compare x=2.08 + 3.01*t* − 1.00*t*^{2} to the first one to get v_{0} and a to plug into the second. v_{0} is 3.01 and 0.5a=-1 so a=-2

make velocity expression v=3.01-2t

Use x=2.08 + 3.01*t* − 1.00*t*^{2} for part a, v=3.01-2t for part b, and since acceleration is constant it has to be -2 for all values of t

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Brenda D.

10/07/19