Question

a spacecraft in deep space fires its thruster, which ejects fuel from the spacecraft. the ejected fuel exerts a constant force on the spacecraft and the total mass of the spacecraft decreases at a constant rate. which of the following graphs best represents the acceleration of the spacecraft as a function of time while the thruster is firing?

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. We can rewrite it as \(a=\frac{F}{m}\).

Step2: Analyze the change of mass with time

The mass \(m\) of the spacecraft decreases at a constant rate. Let the initial mass be \(m_0\) and the rate of mass decrease be \(k\) (a constant), so \(m = m_0-kt\).

Step3: Analyze the acceleration - time relationship

Since \(F\) is constant (given that the ejected fuel exerts a constant force), and \(a=\frac{F}{m}=\frac{F}{m_0 - kt}\). As \(t\) increases, \(m_0-kt\) decreases, and \(a\) increases. The function \(a(t)=\frac{F}{m_0 - kt}\) is a hyperbolic - type function. When \(t = 0\), \(a=\frac{F}{m_0}\). As \(t\) gets larger, the rate of increase of \(a\) with respect to \(t\) slows down. For example, if we consider the derivative of \(a(t)\) with respect to \(t\) using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = F\) (\(u^\prime=0\)) and \(v=m_0 - kt\) (\(v^\prime=-k\)), then \(a^\prime(t)=\frac{Fk}{(m_0 - kt)^{2}}\). As \(t\) increases, the denominator \((m_0 - kt)^{2}\) increases, so the rate of change of acceleration (the slope of the \(a - t\) curve) decreases.

Answer:

A.