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1 year ago

since there are 5,280 feet in one mile and 3.3 feet in one meter how many meters per second are equivalent to 50 miles per hour

Mathematics

1 answer:

Ostrovityanka [42]1 year ago

3 0

Given:

1 mile = 5280 feet

1 meter = 3.3 feet

convert 60 miles per hour to meters per second

60 miles to meters

60 miles x 5280ft x 1meter = (60*5280m)/3.3 = 316800m/3.3

1 mile 3.3ft

= 96,000 meters

1 hour to seconds

1 hour x 60 minutes x 60 seconds = 1 * 60 * 60sec = 3600 seconds

1 hour 1 minute

unit rate: meters per second

96000 meters ÷ 3600 seconds = 26 2400/3600

simplify 26 2400/3600 ⇒ 26 2/3 meters per second

2400 ÷ 1200 = 2

3600 ÷ 1200 = 3

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Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

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