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belka [17]
3 years ago
6

An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,

Physics
2 answers:
Ray Of Light [21]3 years ago
7 0

Explanation:

x = 3.00t^{2} – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = \frac{Total distance}{ Total time}

Average speed = \frac{x (t=2) + x (t=3)}{3}

Average speed = \frac{24+11}{3}

Average speed = \frac{35}{3}

Average speed = 11.66 \frac{m}{s}

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt}

Instantaneous speed(v) = 6t - 2\left \{ {{t=2} \atop {t=3}} \right.

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 \frac{m}{s}

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = \frac{average velocity}{time}

average acceleration =  \frac{11.66}{3-2}

average acceleration = 11.66 \frac{m}{s^{2} }

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = \frac{dv}{dt}

instantaneous acceleration =6

instantaneous acceleration = 6 \frac{m}{s^{2} }

e) for x =0

0 = 3.00t^{2} – 2.00t + 3.00

a = 3, b=-2, c=3

t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}

t= \frac{2 \pm \sqrt{4 - 36} }{6}

t= \frac{2 \pm \sqrt{-32} }{6}

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

Aleks [24]3 years ago
5 0

a) the average speed between t = 2.00 s and t = 3.00 s,   v = 11.66  m/s

b) the instantaneous speed at t = 2.00 s and t = 3.00 s, v = 6  m/s

c) the average acceleration between t = 2.00 s and t = 3.00 s,  v = 11.66  m/s

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.  v = 6  m/s

e) At what time is the object at rest? the given object is not at rest.

<h3>Explanation: </h3>

An object moves along the x-axis according to the equation

x = 3.00t^2 - 2.00t + 3.00

where x is in meters and t is in seconds. Determine :

a) the average speed between t = 2.00 s and t = 3.00 s,

The instantaneous speed or speed is the object speed at a certain instant of time

Average speed = \frac{total distance}{ total time} = \frac{x(t=2s)+x(t=3s)}{3} \\

=\frac{(3.00*2^{2} - 2.00*2 + 3.00)+(3.00*3^{2} - 2.00*3 + 3.00)}{3} \\ =\frac{24+11}{3} \\ =\frac{35}{3} = 11.6 m/s

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt} = 6t - 2

6t - 2 where t=2 and t=3

Instantaneous speed(v) = v(t=2) - v(t=3) = 18-2- (12-2) = 6 m/s

c) the average acceleration between t = 2.00 s and t = 3.00 s, and

The instantaneous acceleration, or acceleration is the limit of the average acceleration when the interval of time approaches 0

average acceleration= \frac{average velocity}{time} =  \frac{11.66}{3-2} =  11.66 m/s

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.

instantaneous acceleration = \frac{dv}{dt} = 6t

6t where t=2 and t=3

Instantaneous acceleration (a) a(t=2 to t=3) = -( 6(2) -  6(3) )= -12 + 18 = 6 m/s^2

e) At what time is the object at rest?

at rest t=0  for x =0

0 = 3.00t^{2} - 2.00t + 3.00 using abc rules,

a = 3, b=-2, c=3

t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}

t= \frac{2 \pm \sqrt{4 - 36} }{6}

t= \frac{2 \pm \sqrt{-32} }{6}

The given object is not at rest because the general solution of this equation gives imaginary value

Learn more about  the instantaneous speed brainly.com/question/11686662

#LearnWithBrainly

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Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

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