Question

Calc a rocket starts from rest and moves upw ce of the earth. for the first 10.0 s of its motion, eration of the rocket is given by a y =( 2.8m /(s³) lirection is upward.

what is the height of t e surface of the earth at t = 10 s?

Answer 1

The height of the rocket from the surface of the earth at t = 10 s is calculated by integrating the given acceleration function to find the velocity, and then integrating the velocity to find the position function. Substituting t = 10 s into the position function, we find that the height of the rocket is 470 m above the earth's surface.

Step-by-step explanation:

To determine the height of the rocket from the surface of the earth at t = 10 s, we should integrate the acceleration function to obtain the velocity function and then integrate the velocity function to get the position function.

Since the acceleration of the rocket is given by a variable acceleration equation a(t) = 2.8 m/s3 * t, the velocity as a function of time, v(t), can be found by integrating the acceleration:

v(t) = ∫ a(t) dt = ∫ (2.8 * t) dt = 2.8 * t2/2 + C

Since the rocket starts from rest, the initial velocity v(0) is 0, which makes the constant of integration (C) equal to 0. Thus, v(t) = 1.4 * t2.

Now, we can find the position as a function of time, s(t), by integrating the velocity function:

s(t) = ∫ v(t) dt = ∫ (1.4 * t2) dt = 1.4 * t3/3 + D

Again, since the rocket starts from a position of 0 at t=0, the constant D is 0, leaving us with s(t) = 0.47 * t3.

To find the position at t = 10 s, we plug the value into our position function:

Height at t = 10 s is s(10) = 0.47 * 103 = 470 m.