Answer:
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.
Explanation:
Credit goes to @Branta
