During the initial stage of launching a spacecraft vertically, the acceleration a(in m/s^2) is a=6t^2. Find the velocity in the spacecraft after 6 seconds.
a = dv/dt, so
v(t) = 2t^3 = 2*6^3
To find the velocity of the spacecraft after 6 seconds, you need to integrate the acceleration function with respect to time.
Given that the acceleration function is a = 6t^2, we need to integrate this function to find the velocity function.
The integral of 6t^2 with respect to t is:
∫6t^2 dt = 2t^3 + C
To find the velocity at t = 6 seconds, we substitute t = 6 into the velocity function:
v(6) = 2(6)^3 + C
Now, we need to determine the constant of integration, C. We can do this by using the initial condition for velocity. Since the spacecraft starts from rest, we know that v(0) = 0.
Substituting t = 0 into our velocity function, we get:
v(0) = 2(0)^3 + C
0 = C
Therefore, C = 0.
Now, we can find the velocity at t = 6 seconds:
v(6) = 2(6)^3 + 0 = 2(216) = 432 m/s
Therefore, the velocity of the spacecraft after 6 seconds is 432 m/s.
To find the velocity of the spacecraft after 6 seconds, we need to integrate the given acceleration function with respect to time.
The given acceleration function is a = 6t^2.
To integrate this function, we use the power rule of integration. The power rule states that ∫x^n dx = (1/(n+1))x^(n+1) + C, where C is the constant of integration.
In our case, since the function is 6t^2, we have n = 2. Applying the power rule, we get:
∫6t^2 dt = (6/(2+1))t^(2+1) + C
= 2t^3 + C
We can now find the velocity by evaluating this integral at t = 6 seconds.
v = 2(6)^3 + C
= 432 + C
Since we are only interested in finding the velocity after 6 seconds, we don't know the constant of integration, C. Therefore, we cannot determine the exact velocity value. However, we know that the velocity after 6 seconds will be 432 units plus some constant value.
Hence, the velocity in the spacecraft after 6 seconds is given by 432 + C.