IB Math AI HL Paper 3 Question Bank

1.) A rocket is launched vertically upward from the ground. The rocket’s height above the ground, y meters, at time t seconds, is given by the following differential equation:
dy/dt = v – 9.8t
where v is the rocket’s initial velocity in meters per second.

a)
i)The initial velocity of the rocket is 50 m/s. Use Euler’s method with a step size of 1 second to estimate the height of the rocket after 5 seconds. Show your calculations.
Euler’s Method Calculation:
We are given the differential equation: dy/dt = v – 9.8t
Using Euler’s method with a step size of 1 second:
t₀ = 0, y₀ = 0 (initial condition)
t₁ = t₀ + h = 0 + 1 = 1
Slope at (t₀, y₀):
m = v – 9.8t_0 = 50 – 9.8(0) = 50
Approximation for y1:
y₁ = y₀ + m * h = 0 + 50 * 1 = 50
Therefore, the estimated height of the rocket after 5 seconds using Euler’s method is 50 meters.

ii)Find the particular solution of the differential equation given the initial condition y(0) = 0.
Finding the Particular Solution:
We are given the differential equation: dy/dt = v – 9.8t
Rearranging the equation, we have:
dy – (v – 9.8t) dt = 0
Integrating both sides:
∫(dy – (v – 9.8t) dt) = ∫0
Integrating, we get:
y – (v – 9.8t) * t/2 = C
Applying the initial condition y(0) = 0:
0 – (v – 9.8(0)) * 0/2 = C
C = 0
Therefore, the particular solution of the differential equation is y – (v – 9.8t) * t/2 = 0.

b)The exact solution of the differential equation is given by y = (1/2)vt – (4.9)t^2. Determine the maximum height reached by the rocket and the time it takes to reach that height. Show your calculations.
Calculating the Maximum Height and Time:
The exact solution of the differential equation is given by y = (1/2)vt – (4.9)t2.
To find the maximum height, we need to determine the time when the velocity, dy/dt, is equal to zero.
Differentiating y with respect to t:
dy/dt = (1/2)v – 9.8t
Setting dy/dt = 0:
(1/2)v – 9.8t = 0
(1/2)v = 9.8t
t = (1/2v) * 9.8
Substituting this value of t back into the equation for y:
y = (1/2)v * ((1/2v) * 9.8) – (4.9) * ((1/2v) * 9.8)2
Simplifying:
y = (1/4) * 9.8 – (4.9) * ((1/4v)2 * 9.82)
y = (1/4) * 9.8 – (4.9) * (1/16v2) * 9.82
y = 2.45 – (0.30625/v2)
The maximum height reached by the rocket is 2.45 meters, and the time it takes to reach that height is (1/2v) * 9.8 seconds.

c) Given that the rocket’s engine can provide a constant thrust of 5000 newtons, find the total work done by the engine from the time of launch until the rocket reaches its maximum height. Show your calculations.
Calculating the Total Work Done by the Engine:
Given that the rocket’s engine can provide a constant thrust of 5000 newtons, we need to calculate the total work done by the engine from the time of launch until the rocket reaches its maximum height.
The work done by the engine is given by the integral of the force exerted by the engine with respect to displacement. In this case, the force exerted by the engine is constant at 5000 newtons.
The displacement of the rocket is equal to the height reached, which is 2.45 meters. Therefore, we can calculate the total work done as follows:
Work = Force × Displacement
Work = 5000 × 2.45
Work = 12250 joules
Therefore, the total work done by the engine from the time of launch until the rocket reaches its maximum height is 12250 joules.

d) Determine the time it takes for the rocket to reach the ground. Show your calculations.
Determine the time it takes for the rocket to reach the ground:
To find the time it takes for the rocket to reach the ground, we need to determine the value of t when the height y is equal to zero.
We have the differential equation:
dy/dt = v – 9.8t
Setting y = 0, we can solve for t:
0 = v – 9.8t
9.8t = v
t = v / 9.8
Therefore, the time it takes for the rocket to reach the ground is t = v / 9.8 seconds.

e) Suppose the rocket’s initial velocity v is such that it can just reach a maximum height of 1000 meters. Determine the range of possible values for v. Show your calculations.
The maximum height reached by the rocket is given by the equation y = (1/2)vt – (4.9)t².
Setting y = 1000, we can solve for v:
1000 = (1/2)v * t – (4.9)t²
Rearranging the equation:
4.9t² – (1/2)vt + 1000 = 0
Using the quadratic formula:
t = (-b ± √(b² – 4ac)) / (2a)
Where a = 4.9, b = -(1/2)v, and c = 1000.
The rocket reaches a maximum height at t > 0, so we only consider the positive root. The rocket reaches its maximum height when the discriminant is equal to zero (b² – 4ac = 0), which gives us a single solution for t.
Solving b2 – 4ac = 0:
(-(1/2)v)² – 4 * 4.9 * 1000 = 0
(1/4)v2 – 4 * 4.9 * 1000 = 0
Solving for v:
v² = 4 * 4.9 * 1000 * 4
v² = 4 * 4.9 * 4000
v² = 4 * 19,600
v² = 78,400
v = ± √78,400
v ≈ ± 280
Therefore, the range of possible values for v is approximately -280 m/s to +280 m/s.
Note: Negative velocity represents the rocket moving downward, while positive velocity represents the rocket moving upward.

f)
i) The mass of the rocket is 2400 kg.If the rocket’s engine experiences a constant drag force of 2000 newtons in the opposite direction of motion, modify the differential equation to include the drag force.
If the rocket’s engine experiences a constant drag force of 2000 newtons in the opposite direction of motion, modify the differential equation to include the drag force and solve for the rocket’s height as a function of time.
The modified differential equation considering the drag force is:
dy/dt = v – 9.8t – Fd/m
where Fd is the drag force (2000 N) and m is the mass of the rocket (2400 kg).
Substituting the values, the modified differential equation becomes:
dy/dt = v – 9.8t – 2000/2400
Simplifying further:
dy/dt = v – 9.8t – 5/6

ii)Use the solution obtained in g)(i) to determine the new maximum height reached by the rocket and the time it takes to reach that height. Show your calculations.
To determine the maximum height and the time taken to reach it, we need to solve the modified differential equation obtained in f)(ii)
Since the modified differential equation is linear, we can solve it using standard techniques. Integrating both sides, we have:
∫dy = ∫(v – 9.8t – 5/6)dt
Integrating with respect to t, we get:
y = vt – (9.8/2)t² – (5/6)t + C
where C is the constant of integration.
Using the initial condition y(0) = 0, we can find the particular solution. Substituting the initial condition into the equation:
0 = v(0) – (9.8/2)(0)² – (5/6)(0) + C
0 = 0 – 0 – 0 + C
C = 0
Therefore, the particular solution is:
y = vt – (9.8/2)t² – (5/6)t
To determine the maximum height reached by the rocket and the time it takes to reach that height, we need to find the vertex of the parabolic function. The maximum height occurs at the vertex of the parabola, which is given by t = -b/2a, where the quadratic equation is in the form ax² + bx + c.
In this case, a = -4.9/2, b = -5/6, and c = 0. Substituting these values, we can find the time taken to reach the maximum height:
t = -(-5/6)/(2*(-4.9/2))
t = 5/6 * 1/4.9
t ≈ 0.214 seconds
Substituting this time back into the equation, we can find the maximum height:
y = v(0.214) – (9.8/2)(0.214)² – (5/6)(0.214)
y ≈ 0.107v – 0.107(4.9) – 0.214(5/6)
y ≈ 0.107v – 0.527 – 0.179
y ≈ 0.107v – 0.706
Therefore, the new maximum height reached by the rocket is approximately 0.107v – 0.706 meters, and it takes approximately 0.214 seconds to reach that height.