# A rocket starts from rest and moves upward from the surface of the earth. For th

sklicatias 2021-11-18 Answered
A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ${a}_{y}=12.80\frac{m}{{s}^{3}}$ )t, where the + y-direction is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?
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## Expert Answer

Daniel Williams
Answered 2021-11-19 Author has 14 answers

a) Since the acceleration is constant constant, integrate the first kinematics equation:
${v}_{y}={\int }_{0}^{t}{a}_{y}dt={\int }_{0}^{10}2.8tdt=1.4{t}^{2}+{v}_{0}$
$y={\int }_{0}^{t}vdt$
with,${v}_{0}=0$ and ${y}_{0}=0$,
$y={\int }_{0}^{10}1.4{t}^{2}dt=0.467{t}^{3}{\mid }_{0}^{10}=467$ m
b) Substitute y=325 m in the equation:
$y=\int vdt=1.4\int {t}^{2}dt=0.467{t}^{3}$
$325=0.476{t}^{3}$

then substitute this time in the equation:
${v}_{y}={\int }_{0}^{t}{a}_{y}dt={\int }_{0}^{8.86}2.8dt=1.4{t}^{2}\mid {0}^{8.86}=110$ m
Result: a) y=467 m
b) ${v}_{y}=110$ m

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