Respuesta :
Explanation:
x = 3.00[tex]t^{2}[/tex] – 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 = [tex]\frac{Total distance}{ Total time}[/tex]
Average speed = [tex]\frac{x (t=2) + x (t=3)}{3}[/tex]
Average speed = [tex]\frac{24+11}{3}[/tex]
Average speed = [tex]\frac{35}{3}[/tex]
Average speed = 11.66 [tex]\frac{m}{s}[/tex]
b) the instantaneous speed at t = 2.00 s and t = 3.00 s,
Instantaneous speed = [tex]\frac{dx}{dt}[/tex]
Instantaneous speed(v) = 6t - 2[tex]\left \{ {{t=2} \atop {t=3}} \right.[/tex]
Instantaneous speed,v(t=2 to t=3) = 18-2-12+2
Instantaneous speed, v = 6 [tex]\frac{m}{s}[/tex]
c) the average acceleration between t = 2.00 s and t = 3.00 s
average acceleration = [tex]\frac{average velocity}{time}[/tex]
average acceleration = [tex]\frac{11.66}{3-2}[/tex]
average acceleration = 11.66 [tex]\frac{m}{s^{2} }[/tex]
d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s
instantaneous acceleration = [tex]\frac{dv}{dt}[/tex]
instantaneous acceleration =6
instantaneous acceleration = 6 [tex]\frac{m}{s^{2} }[/tex]
e) for x =0
0 = 3.00[tex]t^{2}[/tex] – 2.00t + 3.00
a = 3, b=-2, c=3
t= [tex]\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}[/tex]
t= [tex]\frac{2 \pm \sqrt{4 - 36} }{6}[/tex]
t= [tex]\frac{2 \pm \sqrt{-32} }{6}[/tex]
general solution of this equation gives imaginary value. Hence, the given object is not at rest.
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.
Explanation:
An object moves along the x-axis according to the equation
[tex]x = 3.00t^2 - 2.00t + 3.00[/tex]
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 = [tex]\frac{total distance}{ total time} = \frac{x(t=2s)+x(t=3s)}{3} \\[/tex]
[tex]=\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[/tex]
b) the instantaneous speed at t = 2.00 s and t = 3.00 s,
Instantaneous speed [tex]= \frac{dx}{dt} = 6t - 2[/tex]
[tex]6t - 2[/tex] where [tex]t=2[/tex] and [tex]t=3[/tex]
Instantaneous speed(v) [tex]= v(t=2) - v(t=3) = 18-2- (12-2) = 6 m/s[/tex]
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[tex]= \frac{average velocity}{time} = \frac{11.66}{3-2} = 11.66 m/s[/tex]
d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.
instantaneous acceleration [tex]= \frac{dv}{dt} = 6t[/tex]
[tex]6t[/tex] where [tex]t=2[/tex] and [tex]t=3[/tex]
Instantaneous acceleration (a) [tex]a(t=2 to t=3) = -( 6(2) - 6(3) )= -12 + 18 = 6 m/s^2[/tex]
e) At what time is the object at rest?
at rest [tex]t=0[/tex] for [tex]x =0[/tex]
[tex]0 = 3.00t^{2} - 2.00t + 3.00[/tex] using abc rules,
[tex]a = 3, b=-2, c=3[/tex]
[tex]t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}[/tex]
[tex]t= \frac{2 \pm \sqrt{4 - 36} }{6}[/tex]
[tex]t= \frac{2 \pm \sqrt{-32} }{6}[/tex]
The given object is not at rest because the general solution of this equation gives imaginary value
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