The particle travels along the path defined by the parabola&nbsp;y=0.5x2. If the component of velocity along the x axis is vx = (5t) ft/s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.

Question

The particle travels along the path defined by the parabola&amp;nbsp;y=0.5x2. If the component of velocity along the x axis is vx = (5t) ft/s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.

David Tyson · Accepted Answer

From the equation for vx we can get an equation for x:  x=2.5t2  We can insert that to find the equation for y as a function of time:  y=0.5⋅2.52⋅t4=3.125t4  Distance at the t=1 is:  dx=2.5 ft  dy=3.125ft  d=dx+dy=2.52+3.1252  d=4 ft  Acceleration in t=1 is:  ay=37.5t=37.5 fts2  ax=5 fts2  a=ax+ay=52+37.52  a=37.83 fts2  Result: d=4 ft  a=37.83 fts2

fudzisako · Accepted Answer

Answer:403.11ft/s2Explanation:First, let&#039;s find the particle&#039;s position (x, y) at t = 1 s.Given the parabolic path defined by y=0.5x2, we can substitute the x-coordinate into the equation to find the corresponding y-coordinate.When t = 1 s, the x-coordinate is given by x=∫01vxdt, which becomes x=∫015tdt. Integrating, we get x=[52t2]01=52(12−02)=52 ft.Substituting x = 52 into the parabolic equation y=0.5x2, we find y=0.5(52)2=258 ft.Therefore, at t = 1 s, the particle&#039;s position is (x, y) = (52,258) ft.Next, let&#039;s find the magnitude of the particle&#039;s acceleration at t = 1 s.The velocity vector is given by v→=vxi^+vyj^, where i^ and j^ are the unit vectors along the x and y axes, respectively.Given vx=5t ft/s, we can differentiate with respect to time to find the x-component of acceleration ax=dvxdt=5 ft/s^2 (constant).The y-component of acceleration ay can be found by differentiating the parabolic equation y=0.5x2 with respect to time.ay=d2ydt2=d2ydx2·dxdt·(dxdt)Since dydx=dydt·dtdx, we have d2ydx2=ddx(dydx)=ddt(dydx)·dtdx.Differentiating y=0.5x2 with respect to t, we get dydt=ddt(0.5x2)=dxdt·x=5tx.Differentiating dydx with respect to t, we have ddt(dydx)=ddt(5tx)=5x·dxdt+5t·dxdt=10tx+5t2.Substituting these results into ay=d2ydx2·dxdt·(&amp;lt;br&amp;gt;fracdxdt), we obtainay=(10tx+5t2)·5t·5t=250t3x+125t4 ft/s^2.At t = 1 s and x = 52 ft, we can calculate the magnitude of the acceleration as follows:a=ax2+ay2=52+(250(1)3(52)+125(1)4)2=25+(6252+125)2≈403.11 ft/s^2.Therefore, at t = 1 s, the particle&#039;s distance from the origin is x2+y2=(52)2+(258)2=254+62564=625+10064=72564≈3.37 ft, and the magnitude of its acceleration is approximately 403.11ft/s2.

Jazz Frenia · Accepted Answer

Step 1: Given that the particle travels along the path defined by the parabola y=0.5x2, we can relate the position vector r = x𝐢+y𝐣 to the equation of the parabola.Let&#039;s first express y in terms of x. We have:y=0.5x2Next, we need to find x in terms of t using the given component of velocity along the x-axis:vx=5t ft/sSince velocity is the derivative of position with respect to time, we can write:vx=dxdt=5tIntegrating both sides with respect to t:∫dx=∫5tdtx=52t2+CStep 2: Now, let&#039;s determine the constant C. Given that when t = 0, x = 0, and y = 0, we can substitute these values into the equation:0=52(0)2+C0=CTherefore, C = 0, and the equation becomes:x=52t2Now, we can substitute this expression for x into the equation for y:y=0.5x2y=0.5(52t2)2y=258t4Thus, the position vector r of the particle at time t is given by:r=52t2𝐢+258t4𝐣To determine the distance from the origin O, we can calculate the magnitude of the position vector:|r|=(52t2)2+(258t4)2Substituting t = 1 s into the equation, we have:|r|=(52(1)2)2+(258(1)4)2|r|=(52)2+(258)2|r|=254+62564|r|=160064+62564|r|=222564|r|=5898Thus, the particle&#039;s distance from the origin O when t = 1 s is 5898 ft.Step 3:To determine the magnitude of the particle&#039;s acceleration, we can differentiate the velocity with respect to time:a=dvxdt=d2xdt2Differentiating the equation for x:a=d2dt2&amp;lt;br&amp;gt;(52t2)a=52·2ta=5tSubstituting t = 1 s into the equation, we have:a=5(1)a=5 ft/s2Therefore, when t = 1 s, the magnitude of the particle&#039;s acceleration is 5 ft/s2.

The particle travels along the path defined by the parabola y=0.5x^2. If t

Answered question

Answer & Explanation

New Questions in Algebra I