Number 6 c) Increased accuracy canestimate the car's speed after 0.5 secondtraveled.9.4030v(m/s) 20106.time (seconds)Estimate the total distance the object traveled between t = 0 and t=6.6.The velocity of a car is given byv(1) =1² + 2twhere v(t) is measured in ft/sec and t is measured in seconds.a) Estimate the distance traveled from t =1 to t=4 using 3 subdivisions.10b) How accurate is your answer?A car is moving along a straight road and its speed is continually increasing. Spetreadings are recorded at two second intervals and the results are tabulated below.(7.Time (sec)Velocity (ft/sec)4366.8.30384044a) In the first two second interval what is the minimum distance the car could havtraveled? What is the maximum distance?b) Make a lower and upper estimate of the distance the car could have traveled isecond interval.c) If you had to guess how far the car went in the 8 second interval, what wouguess? What is the maximum possible difference between your guess and odistance?d) If the speedometer readings became available for each second, by how muupper estimate exceed the lower estimate?

Name: Number 6 c) Increased accuracy canestimate the car's speed after 0.5 s
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Description: Number 6 c) Increased accuracy canestimate the car's speed after 0.5 secondtraveled.9.4030v(m/s) 20106.time (seconds)Estimate the total distance the object traveled between t = 0 and t=6.6.The velocity of a car is given byv(1) =1² + 2twhere v(t) is m

Answered: Number 6

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Number 6

c) Increased accuracy can
estimate the car's speed after 0.5 second
traveled.
9.
40
30
v(m/s) 20
10
6.
time (seconds)
Estimate the total distance the object traveled between t = 0 and t=6.
6.
The velocity of a car is given by
v(1) =1² + 2t
where v(t) is measured in ft/sec and t is measured in seconds.
a) Estimate the distance traveled from t =1 to t=4 using 3 subdivisions.
10
b) How accurate is your answer?
A car is moving along a straight road and its speed is continually increasing. Spet
readings are recorded at two second intervals and the results are tabulated below.
(7.
Time (sec)
Velocity (ft/sec)
4
36
6.
8.
30
38
40
44
a) In the first two second interval what is the minimum distance the car could hav
traveled? What is the maximum distance?
b) Make a lower and upper estimate of the distance the car could have traveled i
second interval.
c) If you had to guess how far the car went in the 8 second interval, what wou
guess? What is the maximum possible difference between your guess and o
distance?
d) If the speedometer readings became available for each second, by how mu
upper estimate exceed the lower estimate?

Expert Solution

Step 1

$(6)$ The velocity of a car is given by :

$v (t) = t^{2} + 2 t$

here, $v (t)$ is measured in $f t / s e c$ and $t$ is measured in $s e c .$

$(a)$ To determine the distance travelled from $t = 1$ to $t = 4$ using $3$ subdivisions,

$\begin{array}{rcl} \int_{1}^{4} v (t) & = & \int_{1}^{2} (t^{2} + 2 t) d t + \int_{2}^{3} (t^{2} + 2 t) d t + \int_{3}^{4} (t^{2} + 2 t) d t \\ = & {(\frac{t^{3}}{3} + 2 \frac{t^{2}}{2})}_{1}^{2} + {(\frac{t^{3}}{3} + 2 \frac{t^{2}}{2})}_{2}^{3} + {(\frac{t^{3}}{3} + 2 \frac{t^{2}}{2})}_{3}^{4} \\ = & [(\frac{2^{3}}{3} + 2^{2}) - (\frac{1^{3}}{3} + 1^{2})] + [(\frac{3^{3}}{3} + 3^{2}) - (\frac{2^{3}}{3} + 2^{2})] + [(\frac{4^{3}}{3} + 4^{2}) - (\frac{3^{3}}{3} + 3^{2})] \\ = & [(\frac{8}{3} + 4) - (\frac{1}{3} + 1)] + [(9 + 9) - (\frac{8}{3} + 4)] + [(\frac{64}{3} + 16) - (9 + 9)] \\ = & [(\frac{20}{3} - \frac{4}{3})] + [(18 - \frac{20}{3})] + [(\frac{112}{3} - 18)] \\ = & \frac{16}{3} + \frac{34}{3} + \frac{58}{3} \\ = & \frac{108}{3} \\ = & 36 f t . \end{array}$

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