Question
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.)
Ask by Young Boyd. in the United States
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The object is moving in the positive direction on the interval [0, 1.5).
Solution
To determine when the object is moving in the positive direction, we need the velocity to be greater than zero.
The velocity function is given by:
v(t) = 6 – 4t
Set up the inequality for positive velocity:
6 – 4t > 0
Solve the inequality:
6 > 4t
t < 6/4
t < 1.5
Since t is in the interval [0, 3] and the velocity is positive when t < 1.5, the object is moving in the positive direction on the interval [0, 1.5).
Thus, the answer in interval notation is:
[0, 1.5)
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Simplify this solution Mind Expander
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.
