Suppose an object moves along a line with velocity (in \( \mathrm{ft} / \mathrm{s} \) ) \( \mathrm{v}(\mathrm{t})=6-4 \mathrm{t} \), for \( 0 \leq \mathrm{t} \leq 3 \), where t is measured in seconds. The motion is in the positive direction on the interval \( \square \). (Type your answer in interval notation.)

The object’s velocity function is given as \( v(t) = 6 - 4t \). To determine the interval where the object moves in the positive direction, we need to find where \( v(t) > 0 \). Setting up the inequality: \[ 6 - 4t > 0 \] Solving for \( t \): \[ 6 > 4t \] \[ \frac{6}{4} > t \] \[ t < 1.5 \] Therefore, the object moves in the positive direction from \( t = 0 \) to \( t = 1.5 \). In interval notation, the answer is: \[ [0, 1.5) \] In this interval, the velocity is positive, indicating that the object is moving in a positive direction.