Respuesta :
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
Answer:
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
