The acceleration of a certain rocket is given by ax=bt where b is positive constant, find the position and velocity at t=5.0s if Xo=0 and Vo=0 and b=3.0m/s^3?
To find the position and velocity of the rocket at t=5.0s, we need to integrate the acceleration function and apply the initial conditions.
Given:
ax = bt
b = 3.0 m/s^3
t = 5.0 s
Xo = 0 (initial position)
Vo = 0 (initial velocity)
To find the velocity, we integrate the acceleration function with respect to time:
vx = ∫ax dt = ∫bt dt
Integrating bt with respect to t gives:
vx = (1/2)bt^2 + C1
Next, we apply the initial condition Vo = 0 at t = 0:
Vo = vx(t=0) = (1/2)b(0)^2 + C1 = C1
So, we have:
vx = (1/2)bt^2 + Vo
To find the position, we integrate the velocity function with respect to time:
x = ∫vx dt = ∫[(1/2)bt^2 + Vo] dt
Integrating (1/2)bt^2 with respect to t gives:
x = (1/6)bt^3 + Vo*t + C2
Next, we apply the initial condition Xo = 0 at t = 0:
Xo = x(t=0) = (1/6)b(0)^3 + Vo(0) + C2 = C2
So, we have:
x = (1/6)bt^3 + Vo*t + Xo
Substituting the given values:
b = 3.0 m/s^3
t = 5.0 s
Vo = 0 (initial velocity)
Xo = 0 (initial position)
We have:
vx = (1/2)(3.0 m/s^3)(5.0 s)^2 + 0 = 37.5 m/s
x = (1/6)(3.0 m/s^3)(5.0 s)^3 + 0 + 0 = 62.5 m
Therefore, at t = 5.0s, the velocity of the rocket is 37.5 m/s and the position is 62.5 m.
To find the position and velocity at t = 5.0 s, we need to integrate the given acceleration function ax = bt with respect to time.
Given: ax = bt
We can integrate both sides of the equation with respect to time:
∫ ax dt = ∫ bt dt
Integrating both sides will give us the velocity function:
∫ ax dt = ∫ bt dt
Vx = ∫ bt dt
Integrating bt with respect to t will give us:
Vx = (1/2)bt^2 + C₁
Here, C₁ is the constant of integration.
Since the initial velocity (Vo) is given as 0, at t = 0, Vx = Vo = 0, which means C₁ = 0.
Therefore, the velocity function becomes:
Vx = (1/2)bt^2
To find the position function, we need to integrate the velocity function (Vx) with respect to time.
∫ Vx dt = ∫ (1/2)bt^2 dt
Integrating (1/2)bt^2 with respect to t will be:
Xx = (1/6)bt^3 + C₂
Here, C₂ is the constant of integration.
Since the initial position (Xo) is given as 0, at t = 0, Xx = Xo = 0, which means C₂ = 0.
Therefore, the position function becomes:
Xx = (1/6)bt^3
Now, we can substitute the given values to find the position and velocity at t = 5.0 s.
Given: t = 5.0 s, b = 3.0 m/s^3
Using the position function:
Xx = (1/6)b(5.0)^3
Xx = (1/6) * 3.0 * 125.0
Xx = 62.5 m
Therefore, the position at t = 5.0 s is 62.5 m.
Using the velocity function:
Vx = (1/2)b(5.0)^2
Vx = (1/2) * 3.0 * 25.0
Vx = 37.5 m/s
Therefore, the velocity at t = 5.0 s is 37.5 m/s.
I assume in ax=bt, x is a subscript and not a variable. 'a' will be used from here on.
Since initial conditions are all known, the function X(t) can be found by integrations.
b=3.0 m/s³
a(t) = bt
V(t) = ∫a(t) + C1
= (b/2)t² + C1
since V(0)=Vo=0, C1=0, therefore
V(t) = (b/2)t²
X(t) = ∫V(t) + C2
= (b/6)t³ + C2
Since X(0)=Xo=0, C2=0, therefore
X(t)=(b/6)t³
Calculate X(5).