Determining Distance Traveled Before Returning to Starting Position: A Velocity Analysis

In physics, when an object is moving under a given velocity function, understanding the distance it travels is a crucial aspect. This article provides a step-by-step approach to find the distance covered by an object that moves with a velocity given by v(t) 4 - 3t, where t is time in seconds and v(t) is the velocity in meters/second, until it returns to its original position.

Understanding the Problem and Setting Up the Approach

The first step is to determine when the object returns to its original position. This happens when the net displacement (or the position) of the object is zero. To achieve this, we need to integrate the velocity function to find the displacement over time.

Step 1: Determine the Displacement Function

The velocity of the object is given by:

$$v(t) 4 - 3t$$

Integration of the velocity function with respect to time gives the displacement:

$$s(t) int (4 - 3t) dt 4t - frac{3t^2}{2} C$$

Given the initial position s(0) 0, we find that C 0. Therefore, the displacement function is:

$$s(t) 4t - frac{3t^2}{2}$$

Step 2: Find the Time When the Object Returns to the Original Position

The object returns to the original position when:

$$s(t) 0$$

Substitute the displacement function:

$$4t - frac{3t^2}{2} 0$$

Factoring out t, we get:

$$t(4 - frac{3t}{2}) 0$$

Which gives us two solutions:

t 0, the initial position 4 - frac{3t}{2} 0, solving this gives:

$$4 frac{3t}{2} implies t frac{8}{3} text{seconds}$$

Step 3: Calculate the Maximum Displacement Before Returning to the Original Position

To find the maximum displacement, we need to determine when the velocity is zero:

$$v(t) 4 - 3t 0 implies t frac{4}{3} text{seconds}$$

Calculate the displacement at t frac{4}{3}:

$$sleft(frac{4}{3}right) 4left(frac{4}{3}right) - frac{3}{2}left(frac{4}{3}right)^2$$

Calculate each term:

$$4 times frac{4}{3} frac{16}{3}$$ $$frac{3}{2} times frac{16}{9} frac{48}{18} frac{8}{3}$$

Thus:

$$sleft(frac{4}{3}right) frac{16}{3} - frac{8}{3} frac{8}{3} text{ meters}$$

Since the object moves in one direction before returning to the starting point, the total distance covered is:

$$text{Total Distance} 2 times frac{8}{3} frac{16}{3} text{ meters} approx 5.33 text{ meters}$$

Conclusion

The distance covered by the object before it returns to its original position is $frac{16}{3}$ meters or approximately 5.33 meters. This analysis helps in understanding the motion of an object under a specific velocity function.