We can solve this problem using the kinematic equation that relates velocity, acceleration, and displacement:

$v^2 = v_0^2 + 2a(x - x_0)$

Where:

• $v$ is the final velocity (7.8 m/s),
• $v_0$ is the initial velocity (2.9 m/s),
• $a$ is the acceleration (3.4 m/s²),
• $x_0$ is the initial position (1.6 m),
• $x$ is the final position.

We need to solve for the displacement, $\Delta x = x - x_0$.

Rearranging the equation for $x - x_0$:

$v^2 = v_0^2 + 2a(x - x_0)$ $x - x_0 = \frac{v^2 - v_0^2}{2a}$

Now, plug in the values: $x - x_0 = \frac{(7.8)^2 - (2.9)^2}{2(3.4)}$

Let's calculate this.The displacement of the object before it reaches a velocity of 7.8 m/s is approximately 7.7 meters. Therefore, the object will move 7.7 meters from its initial position.

Would you like further details on this solution, or do you have any questions?

Here are 5 related questions to expand on this:

1. How long does it take for the object to reach a velocity of 7.8 m/s?
2. What is the total distance traveled after 5 seconds?
3. If the acceleration was negative, how would the displacement change?
4. How would the problem change if the object started at rest?
5. What would be the displacement if the final velocity was doubled?

Tip: In problems with constant acceleration, the kinematic equations are invaluable for relating displacement, velocity, time, and acceleration.