Math Problem Statement

Question content area top Part 1 The position of an object moving along a line is given by the function s left parenthesis t right parenthesis equals negative 4 t squared plus 20 t. Find the average velocity of the object over the following intervals.

​(a)​ [1, 7​] ​(b) ​[1, 6​] ​(c)​ [1, 5​] ​(d) ​[1, 1plus​h] where hgreater than0 is any real number. Question content area bottom Part 1 ​(a) The average velocity of the object over the interval​ [1, 7​] is    enter your response here.

Solution

To find the average velocity of the object over a given interval [a,b][a, b], we use the formula:

vavg=s(b)s(a)bav_{\text{avg}} = \frac{s(b) - s(a)}{b - a}

The position function is given as: s(t)=4t2+20ts(t) = -4t^2 + 20t

We calculate each part step by step:


(a) Interval [1,7][1, 7]

  1. Compute s(7)s(7) and s(1)s(1): s(7)=4(7)2+20(7)=4(49)+140=196+140=56s(7) = -4(7)^2 + 20(7) = -4(49) + 140 = -196 + 140 = -56 s(1)=4(1)2+20(1)=4(1)+20=4+20=16s(1) = -4(1)^2 + 20(1) = -4(1) + 20 = -4 + 20 = 16

  2. Substitute into the average velocity formula: vavg=s(7)s(1)71=56166=726=12v_{\text{avg}} = \frac{s(7) - s(1)}{7 - 1} = \frac{-56 - 16}{6} = \frac{-72}{6} = -12

Thus, the average velocity over [1,7][1, 7] is: 12\boxed{-12}


(b) Interval [1,6][1, 6]

  1. Compute s(6)s(6) and s(1)s(1): s(6)=4(6)2+20(6)=4(36)+120=144+120=24s(6) = -4(6)^2 + 20(6) = -4(36) + 120 = -144 + 120 = -24 s(1)=16(already computed above).s(1) = 16 \quad \text{(already computed above).}

  2. Substitute into the average velocity formula: vavg=s(6)s(1)61=24165=405=8v_{\text{avg}} = \frac{s(6) - s(1)}{6 - 1} = \frac{-24 - 16}{5} = \frac{-40}{5} = -8

Thus, the average velocity over [1,6][1, 6] is: 8\boxed{-8}


(c) Interval [1,5][1, 5]

  1. Compute s(5)s(5) and s(1)s(1): s(5)=4(5)2+20(5)=4(25)+100=100+100=0s(5) = -4(5)^2 + 20(5) = -4(25) + 100 = -100 + 100 = 0 s(1)=16(already computed above).s(1) = 16 \quad \text{(already computed above).}

  2. Substitute into the average velocity formula: vavg=s(5)s(1)51=0164=164=4v_{\text{avg}} = \frac{s(5) - s(1)}{5 - 1} = \frac{0 - 16}{4} = \frac{-16}{4} = -4

Thus, the average velocity over [1,5][1, 5] is: 4\boxed{-4}


(d) Interval [1,1+h][1, 1 + h]

  1. Compute s(1+h)s(1 + h) and s(1)s(1): s(1+h)=4(1+h)2+20(1+h)s(1 + h) = -4(1 + h)^2 + 20(1 + h) Expand: s(1+h)=4(1+2h+h2)+20(1+h)=48h4h2+20+20hs(1 + h) = -4(1 + 2h + h^2) + 20(1 + h) = -4 - 8h - 4h^2 + 20 + 20h Simplify: s(1+h)=16+12h4h2s(1 + h) = 16 + 12h - 4h^2 s(1)=16(already computed above).s(1) = 16 \quad \text{(already computed above).}

  2. Substitute into the average velocity formula: vavg=s(1+h)s(1)(1+h)1v_{\text{avg}} = \frac{s(1 + h) - s(1)}{(1 + h) - 1} Simplify: vavg=(16+12h4h2)16h=12h4h2hv_{\text{avg}} = \frac{(16 + 12h - 4h^2) - 16}{h} = \frac{12h - 4h^2}{h} Factor and simplify: vavg=124hv_{\text{avg}} = 12 - 4h

Thus, the average velocity over [1,1+h][1, 1 + h] is: 124h\boxed{12 - 4h}


Summary of Results:

  • (a) 12-12
  • (b) 8-8
  • (c) 4-4
  • (d) 124h12 - 4h

Let me know if you'd like more details or have any further questions! Here are some related questions for exploration:

  1. What is the instantaneous velocity of the object at t=1t = 1?
  2. How does the average velocity behave as h0h \to 0 in part (d)?
  3. Can you determine when the object changes direction based on the position function?
  4. What is the displacement of the object over the interval [1,7][1, 7]?
  5. How would the results change if the position function had a constant term?

Tip: The average velocity formula closely resembles the slope of the secant line; interpreting it geometrically helps build intuition.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Average Velocity
Position Function
Algebra
Functions

Formulas

v_avg = (s(b) - s(a)) / (b - a)

Theorems

Average velocity theorem

Suitable Grade Level

Grades 9-11