Question
Suppose an object moves along a line with velocity (in \( \mathrm{ft} / \mathrm{s} \) ) \( v(\mathrm{t})=6-4 \mathrm{t} \), for \( 0 \leq t \leq 3 \), where t is measured in seconds. The motion is in the positive direction on the interval \( [0,1.5) \) : (Type your answer in interval notation.) The motion is in the negative direction on the interval \( \square \). (Type your answer in interval notation.)
Ask by Fowler Ortega. in the United States
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The motion is in the negative direction on the interval (1.5, 3].
Solution
To determine when the object moves in the negative direction, we need to find when the velocity is negative. The velocity function is given by v(t) = 6 - 4t.
1. Set up the inequality for negative velocity:
6 - 4t < 0
2. Solve for t:
6 - 4t < 0 ⟹ -4t < -6 ⟹ t > 1.5
Since t is defined on the interval [0, 3], the object moves in the negative direction when t is in the interval (1.5, 3].
Thus, the motion is in the positive direction on [0, 1.5) and in the negative direction on (1.5, 3].
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Simplify this solution Extra Insights
To determine when the object is moving in the negative direction, we start with the velocity function \( v(t) = 6 - 4t \). The object moves in the negative direction when \( v(t) < 0 \). Set \( v(t) < 0 \): \[ 6 - 4t < 0 \] Solving this inequality gives: \[ 6 < 4t \\ t > \frac{6}{4} \\ t > 1.5 \] Since the object is constrained within the interval \( [0,3] \), the motion will be in the negative direction on the interval \( (1.5, 3] \). Thus, the answer for the negative motion interval is \( (1.5, 3] \). In summary, the motion is in the positive direction on \( [0, 1.5) \) and in the negative direction on \( (1.5, 3] \).
