A rocket is fired at an angle from the top of a tower of height h0 = 68.6 m . Because of the design of the engines, its position coordinates are of the form x(t)=A+Bt2 and y(t)=C+Dt3, where A, B, C , and D are constants. Furthermore, the acceleration of the rocket 1.10 s after firing is
a⃗ =( 2.30 i^+ 2.20 j^)m/s2.
Take the origin of coordinates to be at the base of the tower.
At the instant after the rocket is fired, what is its velocity?
In x v(t) = 2Bt and a(t) = 2B so B = 1.15
In y v(t) = 3Dt^2 and a(t) = 6Dt making D = .33
Then things go a bit off because the velocity at t=0 is zero in x and y and the question becomes pointless.
To find the rocket's velocity at the instant after it is fired, we need to differentiate its position equations with respect to time.
Given:
x(t) = A + Bt^2
y(t) = C + Dt^3
The velocity of the rocket is found by taking the derivative of the position equations:
vx(t) = d/dt (x(t)) = d/dt (A + Bt^2)
= 2Bt
vy(t) = d/dt (y(t)) = d/dt (C + Dt^3)
= 3Dt^2
Now, we can find the velocity of the rocket at the instant after it is fired. Since this happens when t = 0, we substitute t = 0 into the velocity equations:
vx(0) = 2B(0) = 0
vy(0) = 3D(0)^2 = 0
Therefore, the velocity of the rocket at the instant after it is fired is vx = 0 m/s and vy = 0 m/s.
To find the velocity of the rocket at the instant after it is fired, we need to differentiate the position functions with respect to time.
The given position functions of the rocket are:
x(t) = A + Bt^2
y(t) = C + Dt^3
Differentiating x(t) with respect to time (t), we get:
v_x(t) = d(x)/dt = 0 + 2Bt
v_x(t) = 2Bt
Similarly, differentiating y(t) with respect to time (t), we get:
v_y(t) = d(y)/dt = 0 + 0 + 3Dt^2
v_y(t) = 3Dt^2
Now, to find the velocity of the rocket at the instant after it is fired, we substitute t = 1.10 s into the velocity equations:
v_x(1.10) = 2B(1.10)
v_y(1.10) = 3D(1.10)^2
From the given information, we know that the acceleration of the rocket at t = 1.10 s is:
a⃗ = (2.30 i + 2.20 j) m/s^2
Comparing the acceleration and velocity equations, we can equate the corresponding components:
2.30 = 2B(1.10)
2.20 = 3D(1.10)^2
Now, we have a system of equations with two unknowns (B and D). We can solve these equations simultaneously to find B and D.
Dividing the first equation by 2(1.10):
1.05 = B
Dividing the second equation by 3(1.10)^2:
0.671 = D
So, at the instant after the rocket is fired, its velocity is:
v⃗ = (v_x, v_y) = (2B(1.10), 3D(1.10)^2)
= (2(1.05)(1.10), 3(0.671)(1.10)^2)
= (2.31, 2.46) m/s
Therefore, the velocity of the rocket at the instant after it is fired is approximately 2.31 m/s in the x-direction and 2.46 m/s in the y-direction.