Question

A rocket is launched from the ground. Its acceleration, measured every 5 seconds, is tabulated below. Find the velocity and position of the rocket at t = 40 seconds. Use trapezoidal as well as Simpson's rules. Compare the answers: t 0 5 10 15 20 25 30 35 40 u(t) 40.0 45.25 48.50 51.25 54.35 59.48 61.5 64.3 68.7

Answer 1

Answer:

velocity = 2748 m / s

The position of the rocket is 109.92 km above the ground.

Step-by-step explanation:

Acceleration = Velocity / Time

Acceleration refers to the rate of change of velocity with respect to time.

At t = 40 seconds:

acceleration = a = 68.7 m / s²

So, the velocity will be = 68.7 m / s² × 40 s

Velocity is = 2748 m / s.

Distance covered by rocket in 40 seconds = 2748 m / s × 40 s

Distance = 109,920 m = 109.92 km.

So, the position of the rocket is 109.92 km above the ground.

Therefore, we get that the velocity is 2748 m / s and the position of the rocket is 109.92 km above the ground.

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