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.

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