To find the distance moved by the particle during the first second of its motion, we need to integrate the velocity function with respect to time over the interval from 0 to 1 second.
Given the velocity function: v(t) = -2t + 4 m/s
To find the distance, we integrate the absolute value of the velocity function over the interval [0, 1]:
Distance = ∫[0,1] |v(t)| dt
Since the velocity function (-2t + 4) is negative for t > 2, we need to split the integral into two parts:
Distance = ∫[0,2] (-v(t)) dt + ∫[2,1] v(t) dt
Integrating the first part:
∫[0,2] (-v(t)) dt = ∫[0,2] (2t - 4) dt = [t^2 - 4t] from 0 to 2 = (2^2 - 4*2) - (0 - 0) = (4 - 8) - (0 - 0) = -4
Integrating the second part:
∫[2,1] v(t) dt = ∫[2,1] (-2t + 4) dt = [-t^2 + 4t] from 2 to 1 = (-(1)^2 + 4(1)) - (-(2)^2 + 4(2)) = (-1 + 4) - (-4 + 8) = 3 - 4 = -1
Finally, adding the two integrals together:
Distance = (-4) + (-1) = -5 meters
The distance moved by the particle during the first second of its motion is 5 meters, considering the magnitude of the distance.