   Chapter 2.6, Problem 53E

Chapter
Section
Textbook Problem

Velocity Of a Motorcycle The distance s (in feet) covered by a motorcycle traveling in a straight line and starting from rest in t sec is given by the functions(t) = −0.1t3 + 2t2 + 24t (0 ≤ t ≤ 3)Calculate the motorcycle’s average velocity over the time interval [2, 2 + h] for h = 1, 0.1, 0.01, 0.001, 0.0001, and 0.00001, and use your results to guess at the motorcycle’s instantaneous velocity at t = 2.

To determine

To calculate: Average velocity of the motorcycle by the function s(t)=0.1t3+2t2+24t.

Explanation

The given function is s(t)=0.1t3+2t2+24t and time interval is [t,t+h].

Section 1:

The average velocity of the function s(x) is given as,

s(t+h)s(t)h=0.1(t+h)3+2(t+h)2+24(t+h)(0.1(t)3+2(t)2+24(t))h=0.1(t+h)3+2(t+h)2+24(t+h)+0.1(t)32(t)224(t)h={0.1(t3+3t2h+3th2+h3)+2(t2+2th+h2)+24t+24h+0.1(t)32(t)224(t)}h={0.1t30.3t2h0.3th20.1h3+2t2+4th+2h2+24t+24h+0.1t32t224t}h

Simplify the above equation.

s(t+h)s(t)h=h(0.3t20.3th+0.1h2+4t+2h+24)h=0.3t20.3th+0.1h2+4t+2h+24

Average velocity of motorcycle in the time interval [t,t+h] is,

s(t+h)s(t)h=0.3t20.3th+0.1h2+4t+2h+24.

Substitute 2 for t in the above equation.

s(2+h)s(2)h=0.3t20.3×2h+0.1h2+4×2+2h+24=0.3×(2)20.6h+0.1h2+8+2h+24=1.20.6h+0.1h2+2h+32

Average velocity of motorcycle in the time interval [2,2+h] is,

s(2+h)s(2)h=1.20.6h+0.1h2+2h+32 (1)

Substitute 1 for h in the equation (1) and get the value of average velocity in the interval [2,3].

s(2+1)s(2)1=1.20.6h+0.1h2+2h+32=1.20.6×1+0.1×12+2×1+32=1.20.6+0.1+2+32=32.1

Average velocity of motorcycle is 32.1 feet per second in the interval [2,3].

Section 2:

From section 1, Average velocity of motorcycle in the interval [2,2+h] is,

s(2+h)s(2)h=1.20.6h+0.1h2+2h+32 (1)

Substitute 0.1 for h in the equation (1) and get the value of average velocity. in the interval [2,3].

s(2+0.1)s(2)0.1=1.20.6×0.1+0.1(0.1)2+2(0.1)+32=1.20.06+0.01+0.2+32=30.81399

Average velocity of motorcycle in the interval [2,2

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