1) The value of the constant \(A\) in SI units is \(\boxed{500\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}\).
2) The impulse imparted by engine on the rocket during \(2.00\,{\text{s}}\) to \(3.50\,{\text{s}}\) is \(\boxed{5812.5\,{\text{N}} \cdot {\text{s}}}\).
3) The change in velocity of the rocket during the interval \(2.00\,{\text{s}}\) to \(3.50\,{\text{s}}\) is \(\boxed{2.58\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right.\kern-
ulldelimiterspace} {\text{s}}}}\).
Explanation:
Part (1):
The force acting on the rocket is given as:
\({F_x} = A{t^2}\)
Since the magnitude of force experienced by the rocket is \(781.25\,{\text{N}}\) at the instant \(t = 1.25\,{\text{s}}\), the value of constant \(A\) is calculated as:
\(\begin{aligned}781.25&= A{\left( {1.25} \right)^2}\\ A&= \frac{{781.25}}{{1.5625}}\\&= 500\,{{\text{N}} \mathord{\left/ {\vphantom {{\text{N}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{aligned}\)
Thus, the value of the constant in SI units is \(\boxed{500\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}\).
Part (2):
The impulse imparted by the engine is the amount of force exerted in the small interval of time. The impulse imparted by the engine on the rocket during \(2.00\,{\text{s}}\) to \(3.50\,{\text{s}}\) is given by:
\(I=\int\limits_{t = 2.0}^{3.5}{F \cdot dt}\)
Substitute the expression of the force in the above expression of impulse.
\(\begin{aligned}I&= \int\limits_{t = 2.0}^{3.5} {A{t^2}dt}\\&= A\left[ {\frac{{{t^3}}}{3}} \right]_{2.0}^{3.5}\\&= \frac{{500}}{3}\left[ {{{\left( {3.5}\right)}^3} - {{\left( {2.0} \right)}^3}} \right]\\&= 5812.5\,{\text{N}}\cdot {\text{s}}\\\end{aligned}\)
Thus, the imparted by engine on the rocket during \(2.00\,{\text{s}}\) to \(3.50\,{\text{s}}\) is \(\boxed{5812.5\,{\text{N}} \cdot {\text{s}}}\).
Part (3):
According to the impulse-momentum theorem, the impulse imparted by the engine is equal to the change in momentum of the rocket. It is expressed as:
\(\begin{aligned}I&=\Deltap\hfill\\I&=m\times\Deltav \hfill\\\end{aligned}\)
The mass of rocket is \(2250\,{\text{kg}}\) and the impulse imparted by the engine is \(5812.5\,{\text{N}} \cdot {\text{s}}\).
Substitute the values in the above expression.
\(\begin{aligned} 5812.5 &= 2250\times\Delta v\\\Delta v&= \frac{{5812.5}}{{2250}}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-
ulldelimiterspace} {\text{s}}}\\&= 2.58\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right.\kern-
ulldelimiterspace} {\text{s}}}\\\end{aligned}\)
Thus, the change in velocity of the rocket due to the impulse imparted by the engine is
\(\boxed{2.58\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-
ulldelimiterspace} {\text{s}}}}\)
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Impulse-momentum theorem
Keywords: Impulse imparted, the engine exerts, 1.50s interval, starting at 2.00s, rocket’s velocity change, magnitude of force, 2250 kg, in outer space fires an engine, impulse-momentum theorem.
