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&nbsp;ay=12.80ms3&nbsp;)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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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&amp;nbsp;ay=12.80ms3&amp;nbsp;)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?

Daniel Williams · Accepted Answer

a) Since the acceleration is constant constant, integrate the first kinematics equation: vy=∫0taydt=∫0102.8tdt=1.4t2+v0 y=∫0tvdt with,v0=0 and y0=0, y=∫0101.4t2dt=0.467t3∣010=467 m b) Substitute y=325 m in the equation: y=∫vdt=1.4∫t2dt=0.467t3 325=0.476t3 ∴t=8.86 s then substitute this time in the equation: vy=∫0taydt=∫08.862.8dt=1.4t2∣08.86=110 m Result: a) y=467 m b) vy=110 m

xleb123 · Accepted Answer

(a) Let&#039;s find the velocity function first. We integrate the given acceleration function ay=12.80ms3t with respect to time t:vy=∫aydt=∫12.80ms3tdtIntegrating the expression on the right side, we get:vy=12.80ms3·12t2+C1where C1 is the constant of integration. Since the rocket starts from rest, we know that the initial velocity is zero, so we can substitute t=0 and vy=0 into the equation to solve for C1:0=12.80ms3·12(0)2+C1C1=0Therefore, the velocity function becomes:vy=12.80ms3·12t2Now, let&#039;s find the height function by integrating the velocity function with respect to time:y=∫vydt=∫12.80ms3·12t2dtIntegrating the expression on the right side, we get:y=12.80ms3·16t3+C2where C2 is the constant of integration. To determine the value of C2, we can substitute t=0 and y=0 into the equation:0=12.80ms3·16(0)3+C2C2=0Therefore, the height function becomes:y=12.80ms3·16t3To find the height of the rocket above the surface of the Earth at t=10.0 s, we substitute t=10.0 s into the height function:y=12.80ms3·16(10.0s)3Calculating the expression, we get:y=2133.33mTherefore, the height of the rocket above the surface of the Earth at t=10.0 s is 2133.33 m.(b) To find the speed of the rocket when it is 325 m above the surface of the Earth, we need to find the time t at which the height y is equal to 325 m. We set up the equation as follows:325m=12.80ms3·16t3Simplifying the equation, we have:t3=325m·612.80ms3t3=150Taking the cube root of both sides, we get:t=1503Calculating the expression, we find:t≈5.848sNow that we have the time t, we can substitute it into the velocity function to find the speed of the rocket:vy=12.80ms3·12(5.848s)2Calculating the expression, we get:vy≈202.92msTherefore, the speed of the rocket when it is 325 m above the surface of the Earth is approximately 202.92 m/s.

fudzisako · Accepted Answer

To find the height of the rocket above the surface of the earth at t=10.0 s, we need to integrate the acceleration function ay=12.80ms3t with respect to time t from 0 to 10.0 s to obtain the velocity function vy(t) and then integrate the velocity function with respect to time again to obtain the displacement function y(t).Integrating ay with respect to t, we get:∫0taydt=∫0t12.80ms3tdtUsing the power rule for integration, the integral becomes:∫0taydt=[12.802ms3t2]0tEvaluating the integral limits, we have:vy(t)=12.802ms3t2Integrating vy(t) with respect to t, we get:∫0tvy(t)dt=∫0t12.802ms3t2dtUsing the power rule for integration again, the integral becomes:∫0tvy(t)dt=[12.806ms3t3]0tEvaluating the integral limits, we have:y(t)=12.806ms3t3Substituting t=10.0 s into the displacement function y(t), we find the height of the rocket above the surface of the earth at t=10.0 s:y(10.0)=12.806ms3(10.03)Simplifying the expression, we can calculate the height.

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

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