Answer:
To find the velocity after one second and the distance traveled by the object in the first two seconds, we can integrate the given force function over the respective time intervals.
Velocity after one second (t = 1):
We need to find the integral of the force function from t = 0 to t = 1, and then divide by the mass of the object (2 kg) to get the velocity.
The force function is given as F(t) = 12t - 24 N.
Integrating F(t) with respect to t, we get:
∫F(t) dt = ∫(12t - 24) dt = 6t^2 - 24t + C
To find the constant of integration (C), we can use the initial condition that the object is 5 meters to the right of the origin (t = 0). Plugging in t = 0 and the position x = 5 into the integrated equation, we have:
6(0)^2 - 24(0) + C = 5
C = 5
Therefore, the integrated function becomes:
∫F(t) dt = 6t^2 - 24t + 5
Now, to find the velocity after one second (t = 1), we evaluate the integrated function:
v(1) = (6(1)^2 - 24(1) + 5) / 2
v(1) = (6 - 24 + 5) / 2
v(1) = -13/2
v(1) = -6.5 m/s
So, the velocity after one second is -6.5 m/s.
Distance traveled in the first two seconds (t = 2):
To find the distance traveled, we need to calculate the area under the velocity-time graph. Since the object comes to rest after 2 seconds, the distance traveled will be the integral of the velocity function from t = 0 to t = 2.
The velocity function is v(t) = 6t^2 - 24t + 5.
Integrating v(t) with respect to t, we get:
∫v(t) dt = ∫(6t^2 - 24t + 5) dt = 2t^3 - 12t^2 + 5t + C
Again, we can determine the constant of integration (C) by using the initial condition that the object is 5 meters to the right of the origin (t = 0). Plugging in t = 0 and the position x = 5 into the integrated equation, we have:
2(0)^3 - 12(0)^2 + 5(0) + C = 5
C = 5
Therefore, the integrated function becomes:
∫v(t) dt = 2t^3 - 12t^2 + 5t + 5
Now, to find the distance traveled in the first two seconds (t = 2), we evaluate the integrated function:
Distance = ∫v(t) dt (from t = 0 to t = 2)
Distance = [(2(2)^3 - 12(2)^2 + 5(2) + 5) - (2(0)^3 - 12(0)^2 + 5(0) + 5)]
Distance = (16 - 48 + 10 + 5) - (0 - 0 + 0 + 5)
Distance = -12
So, the distance traveled by the object in the first two seconds is -12 meters. Note that the negative sign indicates that the object
Explanation: