A spaceship is traveling directly away from earth with speed 4t2+6 km/s. At time t=0 is is 250 km from earth. How far is it from earth one minute after time t=0? Include units in your answer.

576610 km

Explanation:

the speed is dx/dt which is 4t² + 6

On integration, we have,

x = 8t³/3 + 6t + C

C is the constant of integration,

t is the time in second,

x is the distance in km

So, at t = 0, and x = 250, we have

250 = 8(0)³/3 + 6 * 0 + C

250 = 0 + 0 + C, so that,

C = 250

x = 8t³/3 + 6t + C, where C = 250

x = 8t³/3 + 6t + 250

at time, t = 1 min = 60 s

x = 8(60³)/3 + 6(60) + 250

x = (8*216000)/3 + 360 + 250

x = 1728000/3 + 610

x = 576000 + 610

x = 576610 km

The spaceship is 288,610km from the earth one minute after t = 0.

L(60) = 288,610 km

Explanation:

Let L represent how far the spaceship is from the earth.

The speed the spaceship can be given as;

dL/dt = 4t^2 + 6

So,

L = ∫ dL/dt = ∫ 4t^2 + 6

L(t) = (4t^3)/3 + 6t + C

At t = 0, L(0) = 250 km

L(0) = (4(0)^3)/3 + 6(0) + C = 250

C = 250

Then,

L(t) = (4t^3)/3 + 6t + 250 1

Since dL/dt is in m/s

time t is in seconds.

One minute after t = 0 is t = 1 minutes = 60 seconds.

Substituting t = 60 into equation 1;

L(60) = (4(60)^3)/3 + 6(60) + 250

L(60) = (4×216000)/3 + 360 + 250

L(60) = 288,610 km

Therefore, the spaceship is 288,610km from the earth one minute after t = 0

the pull of gravity on the rocket decreases

ill do it for 100, also this isnt 80 its 40