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At launch the space shuttle weighs $4.5$ million pounds. When it is launched from rest, it takes $8.00 \mathrm{~s}$ to reach $161 \mathrm{~km} / \mathrm{h}$ , and at the end of the first $1.00 \mathrm{~min}$ its speed is $1610 \mathrm{~km} / \mathrm{h}$ . $(a)$ What is the average acceleration (in $\mathrm{m} / \mathrm{s}^2$ ) of the shuttle $(i)$ during the first $8.00 \mathrm{~s}$ , and $(ii)$ between $8.00 \mathrm{~s}$ and the end of the first $1.00 \mathrm{~min}$ ? $(b)$ Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the shuttle travel $(i)$ during the first $8.00 \mathrm{~s}$ , and $(ii)$ during the interval from $8.00 \mathrm{~s}$ to $1.00 \mathrm{~min}$ ?

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

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Question

At launch a rocket ship weighs $4.5$ million pounds. When it is launched from rest, it takes $8.00$ s to reach $161$ km/h; at the end of the first $1.00$ min, its speed is $1610$ km/h.
(b) Assuming the acceleration is constant during each time inter- val (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first $8.00$ s and (ii) during the interval from $8.00$ s to $1.00$ min?

Solution

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Answered 1 year ago

(b)
This is the analysis model for particle under constant acceleration.
To calculate the distance, we will use the following equation:
$\color{Fuchsia} x_f=x_i+\dfrac{1}{2}(v_{xi}+v_{xf})t$
(i) In this time interval the rocket launched from rest, so $v_{xi}=0$
$x_f= 0+\dfrac{1}{2}(0+44.7) (8)$
$\color{#4257b2} = \boxed{178.8\;\mathrm{m}}$

(ii) Here the time interval is not starting from Zero, so the change in $v_x$ from $44.7\;\mathrm{m/s}$ to $ 447.2;\mathrm{m/s}$ will happen in the time interval $t =60-8=52\;\mathrm{s}$
$x_f= 0+\dfrac{1}{2}(44.7+447.2) (52)$
$\color{#4257b2}=\boxed{1.28\times10^4\;\mathrm{m}}$