Answer: a) To determine when the particle is moving right, we need to find when its velocity is positive. The velocity of the particle is given by the derivative of its position function:
v(t) = 2t^2 - 6t + 4
We can factor this quadratic expression to get:
v(t) = 2(t - 1)(t - 2)
So the velocity is positive for t > 2 and t < 1. Therefore, the particle is moving right during the time interval 1 < t < 2.
b) The velocity function v(t) found in part (a) is a quadratic function with a negative leading coefficient, which means that it is maximized at its vertex. The vertex of the quadratic function is given by:
t = -b / (2a)
where a = 2, b = -6. Plugging in these values, we get:
t = -(-6) / (2 * 2) = 1.5
Therefore, the maximum velocity occurs at t = 1.5 seconds. To find the maximum velocity, we can plug this value of t into the velocity function:
v(1.5) = 2(1.5)^2 - 6(1.5) + 4 = -1
So the maximum velocity is -1 cm/s.
c) The acceleration of the particle is given by the derivative of its velocity function:
a(t) = 4t - 6
This is a linear function with a positive slope, which means that the acceleration is constant and increasing. The maximum acceleration occurs at t = 4 seconds, and its value is:
a(4) = 4(4) - 6 = 10
So the maximum acceleration is 10 cm/s^2.
d) The particle is speeding up when its acceleration is positive. From part (c), we know that the acceleration is positive for all t in the interval 0 ≤ t ≤ 4, so the particle is speeding up during this entire time interval.
Explanation: