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Calculus IChapter 4: Application of DerivativesSection 4-1

Rates of Change — Question 1

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Question 1

The position of an object moving along a straight line is given by: s(t)=3t3−5t2+2t(in meters, where t is in seconds)s(t) = 3t^3 - 5t^2 + 2t \quad \text{(in meters, where $t$ is in seconds)}

(a) Find the velocity function v(t)v(t).

(b) Find the instantaneous velocity at t=2t = 2 seconds.

(c) Find the average velocity between t=1t = 1 and t=2t = 2.

Original worksheet page 1: question and worked solution for 4-1-001
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Question 1 - Solution

We are given: s(t)=3t3−5t2+2ts(t) = 3t^3 - 5t^2 + 2t

(a) Velocity is the derivative of position: v(t)=s′(t)=9t2−10t+2v(t) = s'(t) = 9t^2 - 10t + 2

(b) Instantaneous velocity at t=2t = 2: v(2)=9(4)−10(2)+2=36−20+2=18v(2) = 9(4) - 10(2) + 2 = 36 - 20 + 2 = 18

Answer: v(2)=18m/s\boxed{v(2) = 18 \ \text{m/s}}

(c) Average velocity from t=1t = 1 to t=2t = 2: s(2)−s(1)2−1=s(2)−s(1)\frac{s(2) - s(1)}{2 - 1} = s(2) - s(1)

Compute s(2)s(2): s(2)=3(8)−5(4)+2(2)=24−20+4=8s(2) = 3(8) - 5(4) + 2(2) = 24 - 20 + 4 = 8

Compute s(1)s(1): s(1)=3(1)−5(1)+2(1)=3−5+2=0s(1) = 3(1) - 5(1) + 2(1) = 3 - 5 + 2 = 0

Average velocity=8−01=8\text{Average velocity} = \frac{8 - 0}{1} = 8

Answer: 8m/s\boxed{8 \ \text{m/s}}

Original worksheet page 2: question and worked solution for 4-1-001

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