12. Find the equations for a linear trajectory that starts at (0. 50) and ends a
ID: 2884806 • Letter: 1
Question
12. Find the equations for a linear trajectory that starts at (0. 50) and ends at (200, 350), where the speed isa constant 25 m/s. Be sure to specify the time interval. 13. Find the equations for a linear trajectory of an object that starts at (0, 0) and ends at (300, 400), where the speed increases according to S)- 2t. Begin by assuming the equation of the trajectory has the form ra)-(a+bu(t).c+du(t), where u't)> 0 14. How much time is required for the object in Step 13 to complete the trip? 15. The approach outlined above, which works well for circular and linear trajectories, encounters difficulties or more general trajectories. Suppose an object has a trajectory given by r() (.go)for OsiST where fand g are given differentiable functions. These equations give the shape of the trajectory, but it may not give the desired speed. So assume that a speed function S is also specified that describes an object's motion along the path. Let's follow the above procedure to find the equations of the desired trajectory. As before we introduce the function u), where u'(a)>0, and assume the equation of the trajectory has the form r)(u()).g(u()Use the Chain Rule to show that the speed is given by And now we see the difficulty: The functions f, g, and S are given and the challenge is to solve this equation for u(t). For circular and linear trajectories, this equation is solvable. For general functions f and g it is usually difficult to solve. Here are some other cases in which a solution may be found.Explanation / Answer
(0,0) to (300,400)
This is how it is done :
r(t) = (0,0) + t<(300,400) - (0,0)>
r(t) = <0,0> + t<300,400>
r(t) = <0,0> + <300t,400t>
r(t) = <0 + 300t , 0 + 400t>
r(t) = <300t , 400t> -----> ANS
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14)
Distance = sqrt((300t)^2 + (400t)^2)
= sqrt(900t^2 + 1600t^2)
= sqrt(2500t^2)
= 500t
Speed = 2t
Time = distance/speed
= 500t/(2t)
= 250 seconds -----> ANS
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