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4 years ago

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]4 years ago

7 0

Explanation:

x = 3.00 $t^{2}$ – 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 = 3.00 $t^{2}$ – 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]4 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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Part b

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