X=2+2t+t^2 X0?,V0?,a?, V(t)?, S(t)? V через 5 секунд

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вопрос

Given that X = 2 + 2t + t^2, we can find the initial position X0, initial velocity V0, acceleration a, velocity V(t), and displacement S(t).

Initial position (X0):
At t = 0, X = 2 + 2(0) + (0)^2 = 2.
Therefore, X0 = 2.

Initial velocity (V0):
The initial velocity is the derivative of X with respect to time (t):
V(t) = dX/dt = d/dt(2 + 2t + t^2) = 2 + 2t.
At t = 0, V0 = 2.

Acceleration (a):
The acceleration is the derivative of V with respect to time (t):
a = dV/dt = d/dt(2 + 2t) = 2.
Therefore, the acceleration is constant and equal to 2.

Velocity (V(t)):
V(t) = 2 + 2t.

Displacement (S(t)):
The displacement is the integral of V(t) with respect to time:
S(t) = ∫V(t) dt = ∫(2 + 2t) dt = 2t + t^2 + C.
At t = 0, S(0) = 0 (since X0 = 2).
Thus, the displacement function is S(t) = 2t + t^2.

Velocity after 5 seconds:
V(5) = 2 + 2(5) = 2 + 10 = 12.
Therefore, the velocity after 5 seconds is 12.