Question textConsider a spacecraft approaching a distant asteroid. The spacecraft’s distance d (km) from the asteroid over time t (seconds) can be approximated by the equation: $$ d(t)=807.1758e^{-0.0011t} $$a) Use the bisection method with the initial guesses of 500s and 1000s and a precision of 1e-6 to estimate the time it will take for the spacecraft to reach 300 km from the asteroid.Answer 1 Question 3[input] s (to 2 decimal places)b) The spacecraft needs to activate its landing procedure when it is 70km from the asteroid’s surface. Use the false position method with the initial guesses of 2000s and 2500s and a precision of 1e-6 to estimate the time at which the landing procedure needs to be activated.Answer 2 Question 3[input] s (to 2 decimal places)

Question

Question textConsider a spacecraft approaching a distant asteroid. The spacecraft’s distance d (km) from the asteroid over time t (seconds) can be approximated by the equation: $$ d(t)=807.1758e^{-0.0011t} $$a) Use the bisection method with the initial guesses of 500s and 1000s and a precision of 1e-6 to estimate the time it will take for the spacecraft to reach 300 km from the asteroid.Answer 1 Question 3[input] s (to 2 decimal places)b) The spacecraft needs to activate its landing procedure when it is 70km from the asteroid’s surface. Use the false position method with the initial guesses of 2000s and 2500s and a precision of 1e-6 to estimate the time at which the landing procedure needs to be activated.Answer 2 Question 3[input] s (to 2 decimal places)

BlackTom AI · Accepted Answer

We start by restating the problem to ensure clarity: we have the distance to the asteroid as a function of time, d(t) = 807.1758 e^{-0.0011 t}, and we must find two times using two different root-finding methods, each with specified initial guesses and a tolerance of 1e-6. Option a) Find t when d(t) = 300 km using the bisection method with initial guesses t0 = 500 s and t1 = 1000 s. In a root-finding formulation, we define f(t) = d(t) - 300. The bisection method requires that f(t0) and f(t1) have opposite signs, indicating a root in [t0, t1], and then repeatedly bisects the interval, choosing the subinterval where the sign change occurs, until the interval width is within the desired tolerance (here, 1e-6) or the function values are sufficiently close to zero. The steps involve evaluating f at the midpoint repeatedly, updating either the left or right endpoint based on the sign of f(mid). In this specific setup, f(500) = 807.1758 e^{-0.0011×500} − 300 and f(1000) = 807.1758 e^{-0.0011×1000} − 300. Since e^{-0.55} ≈ 0.577 and e^{-1.1} ≈ 0.333, f(500) ≈ 807.1758×0.577 − 300 ≈ 465.3 − 300 > 0, while f(1000) ≈ 807.1758×0.333 − 300 ≈ 268.8 − 300 < 0. This confirms a root lies between 500 and 1000 seconds, so bisection will converge to the time t where d(t) crosses 300 km. Each iteration halves the interval, and the stated tolerance 1e-6 translates to a very small interval length, yielding a final estimate of t ≈ 899.78 s when rounded to two decimals. The key idea is that the exponential decay makes d(t) strictly decreasing, ensuring a single crossing of 300 km in the interval, so bisection is appropriate. Option a correctness discussion: The transformation f(t) = d(t) − 300 captures the root condition. The initial interval [500, 1000] brackets the root because f(500) > 0 and f(1000) < 0, fulfilling the bisection precondition. The method’s halving behavior systematically narrows the root, and the exponential form guarantees monotonic decay, preventing multiple roots in the interval. The final numeric value aligns with the exact analytic solution t = (−ln(300/807.1758))/0.0011, which is approximately 899.78 s, so the procedure is consistent with the closed-form result. Option b) Find t when d(t) = 70 km using the false position (regula falsi) method with initial guesses t0 = 2000 s and t1 = 2500 s. Here we define f(t) = d(t) − 70. Again, the false position method requires f(t0) and f(t1) to have opposite signs to ensure a root lies in [t0, t1], and then it computes a new point t2 where the line through (t0, f(t0)) and (t1, f(t1)) crosses the t-axis, i.e., t2 = t1 − f(t1) (t1 − t0) / (f(t1) − f(t0)). The iterations continue by evaluating f(t2) and replacing either t0 or t1 with t2 depending on the sign, while preserving a bracketing interval around the root. With d(t) being a strictly decreasing exponential, there will be a unique crossing with 70 km somewhere at a later time, and the method will converge to that root as the tolerance tightens to 1e-6. The calculation at the root would satisfy e^{-0.0011 t} = 70/807.1758, which yields t = −(1/0.0011) ln(70/807.1758). Since 70/807.1758 ≈ 0.0867 and ln(0.0867) ≈ −2.446, the analytic result is t ≈ 2222.77 s, which matches what the false position method would converge to given the specified bounds and tolerance. Option b correctness discussion: The false position method operates on f(t) = d(t) − 70, seeking the zero crossing. The chosen initial interval [2000, 2500] must bracket the root, which is plausible here because d(t) is monotonically decreasing; at t = 2000 s, d(2000) is still above 70 km, and at t = 2500 s, d(2500) has fallen below 70 km, yielding opposite signs for f(t0) and f(t1). The regula falsi update formula uses a weighted secant-like step that preserves a bracketing interval while attempting to accelerate convergence compared to pure bisection. The exact analytic solution, t = (−ln(70/807.1758))/0.0011, gives about 2222.77 s, so the false position method should converge toward that value within the stated tolerance, assuming standard implementation. Summary of the numerical relationships being exploited: the given distance model is an exponential decay, d(t) = A e^{−kt} with A = 807.1758 and k = 0.0011. For a target distance D, the root is t = (ln(A/D))/k. The first target D = 300 yields t ≈ (ln(807.1758/300))/0.0011 ≈ 899.78 s, and the second target D = 70 yields t ≈ (ln(807.1758/70))/0.0011 ≈ 2222.77 s. The place where distance equals 300 or 70 corresponds to the crossing points that the respective methods are designed to locate, starting from their respective initial guesses and driving the error to within 1e-6.

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