Question: Give the position function s = (t) of an object moving along the s-axis as a function of time t. Graph together with the
Give the position function s = ƒ(t) of an object moving along the s-axis as a function of time t. Graph ƒ together with the velocity function y(t) = ds/dt = ƒ′(t) and the acceleration function a(t) = d2s/dt2 = ƒ″(t). Comment on the object’s behavior in relation to the signs and values of v and a. Include in your commentary such topics as the following:
a. When is the object momentarily at rest?
b. When does it move to the left (down) or to the right (up)?
c. When does it change direction?
d. When does it speed up and slow down?
e. When is it moving fastest (highest speed)? Slowest?
f. When is it farthest from the axis origin?
s = 4 - 7t + 6t2 - t3, 0 ≤ t ≤ 4
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To graph the position velocity and acceleration functions first find the velocity and acceleration functions by taking the first and second derivatives of the position function respectively st 4 7t 6t2 t3 vt st 7 12t 3t2 at vt 12 6t To graph these functions we can plot them on a coordinate plane with the yaxis representing the values of st vt or at and the xaxis representing time t a The object is momentarily at rest when its velocity is zero or vt 0 So we need to solve for the value of t that makes vt 0 0 7 12t 3t2 0 3t2 12t 7 Using the quadratic formula t 12 sqrt122 437 23 131 236 Therefore the object is momentarily at rest at t 131 and t 236 b The object moves to the left down when its velocity is negative or vt 0 It moves to the right up when its velocity is positive or vt 0 From the velocity function we can see that vt is negative when t 1 and when t 2 and positive when 1 t 2 c The object changes direction when its velocity changes sign or when vt 0 We found earlier that vt 0 at t 131 and t 236 so the object changes direction at these points d The object speeds up when its acceleration is positive or at 0 It slows down when its acceleration is negative or at 0 From the acceleration function we can see that at is positive when t 2 and negative when t 2 e The object is moving fastest when its speed absolute value of velocity is highest We can find the highest speed by finding the maximum value of vt vt 7 12t 3t2 To find the maximum value we take the derivative ... View full answer
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