A particle moves along the x-axis according to the equation x = 50.0t + 10.0t^2, where x is in meters and t is in seconds. Calculate the average velocity of the particle during the first 2.30 s of its motion.

Assume the particle starts at x=0.

What is the position after 2.3 s?

x = (50 * 2.3) + (10 * 2.3^2)

That's how far it has travelled in 2.3s

From there the average velocity is easy, I think

To calculate the average velocity of the particle during the first 2.30 seconds, we need to find the displacement of the particle and divide it by the time interval.

1. First, we need to find the position of the particle at the starting time, t1 = 0 s.
Substitute t = 0 into the equation x = 50.0t + 10.0t^2:
x1 = 50.0(0) + 10.0(0)^2
x1 = 0

The initial position of the particle, x1, is 0 meters.

2. Next, we need to find the position of the particle after 2.30 seconds, t2 = 2.30 s.
Substitute t = 2.30 into the equation x = 50.0t + 10.0t^2:
x2 = 50.0(2.30) + 10.0(2.30)^2
x2 = 115.0 + 53.38
x2 = 168.38

The final position of the particle, x2, is 168.38 meters.

3. Now, we can calculate the displacement of the particle during the first 2.30 seconds.
Displacement (Δx) = x2 - x1
Δx = 168.38 - 0
Δx = 168.38 meters

4. Lastly, we calculate the average velocity by dividing the displacement by the time interval.
Average velocity = Δx / (t2 - t1)
Average velocity = 168.38 / 2.30
Average velocity ≈ 73.22 m/s

Therefore, the average velocity of the particle during the first 2.30 seconds of its motion is approximately 73.22 m/s.