Answer:
A) At 2.2 s the position of the particle is 9.4 m.
B) At t =2.2 s the velocity is 9.7 m/s.
C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²
Explanation:
Hi there!
A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:
dx/dt = 2 · t²
Separating variables, we can find the equation of position as a function of time:
dx = 2 · t² · dt
Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:
∫ dx = 2 ∫ t² · dt
x - 2.3 m = 2/3 · t³
x = 2.3 m + 2/3 m/s³ · t³
Replacing t = 2.2 s:
x = 2.3 m + 2/3 m/s³ · (2.2 s)³
x = 9.4 m
At 2.2 s the position of the particle is 9.4 m
B) Now, let´s evaluate the velocity function at t = 2.2 s:
v = 2 · t²
v = 2 m/s³ · (2.2 s)²
v = 9.7 m/s
At t =2.2 s the velocity is 9.68 m/s
C) The acceleration is the variation of the velocity over time (the derivative of the velocity):
dv/dt = a
a = 4 · t
At t = 2.2 s:
a = 4 m/s³ · 2.2 s
a = 8.8 m/s²
At t = 2.2 s the acceleration of the particle is 8.8 m/s²
(A) The particle's position at time, t = 2.2 s is 7.1 m.
(B) The velocity of the particle at 2.2 s is 9.68 m/s.
(C) The acceleration of the particle at 2.2 s is 8.8 m/s².
The given parameters:
The particle's position at time, t = 2.2 s is calculated as follows;
[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]
The velocity of the particle at 2.2 s is calculated as follows;
[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]
The acceleration of the particle at 2.2 s is calculated as follows;
[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]
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