Question

Professor Hines is performing a nuclear experiment in a particle accelerator. The particle has an initial position of -24 meters, an initial velocity of 1 m/s, and is subjected to a constant acceleration of 292 m/s/s for 1.1 seconds. All motion is in a straight line.
***I found the velocity, now I need the average velocity: What is the average velocity at the end of the time interval?

Answer 1

Have you considered the formulation for Uniform Acceleration?

Acceleration: a(t) = 292
Velocity: v'(t) = a(t), thus v(t) = 292t + 1
Position: x'(t) = v(t), thus x(t) = 146t^2 + t - 24

We didn't really need x(t) for this problem. What is v(1.1)?

Answer 2

I just solved it, v2=v1+acceleration(time)
1m/s+ 292(1.1) = 322.2 m/s

Answer 3

Correct. v = u + at = 1 + 292(1.1) = 322.2 m/s

Answer 4

a(1.1) = 292*1.1 + 1 = 322.2 -- There you go.