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All Calculus AB Resources

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Flashcards: Connect Position, Velocity, and Acceleration

A particle is moving in a straight path with a constant initial velocity. The particle is then subjected to a force causing a time-dependent acceleration given as a function of time:

a(t)=(a+b)t

After 10 seconds, the particle has a velocity equal to meters-per-second. Find the initial velocity in terms of the constants , and

Units are all in S.I. (meters, seconds, meters-per-second, etc.)

v_o =k-50(a+b)

$v_o =\frac{b}{a}+k$

v_o =b-5(a+k)

$v_o =\frac{k}{a+b}-50a$

$v_o =\frac{k}{a}+b$

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Flashcards: Connect Position, Velocity, and Acceleration

A particle is moving in a straight path with a constant initial velocity. The particle is then subjected to a force causing a time-dependent acceleration given as a function of time:

a(t)=(a+b)t

After 10 seconds, the particle has a velocity equal to meters-per-second. Find the initial velocity in terms of the constants , and

Units are all in S.I. (meters, seconds, meters-per-second, etc.)

Your answer:

$v_o =\frac{b}{a}+k$

Your answer:

v_o =b-5(a+k)

Your answer:

$v_o =\frac{k}{a+b}-50a$

Your answer:

$v_o =\frac{k}{a}+b$

Correct answer:

v_o =k-50(a+b)

Explanation:

a(t)=(a+b)t

Begin by finding the velocity function by integrating the acceleration function.

$v(t)=\int a(t)dt = \int (a+b)tdt=\frac{1}{2}(a+b)t^2+v_o$

We use v_o as the constant of integration since the function v(t) is a velocity and at initially, at t = 0 , the velocity is equals the constant of integration.

$v(t)=\frac{1}{2}(a+b)t^2+v_o$

At t = 10 seconds we are told the velocity is equal to .
$v(t=10)=k\: \: \: \: \: \:\rightarrow\: \: \: \: \: \:\frac{1}{2}(a+b)100+v_o =k$

50(a+b)+v_o =k

v_o =k-50(a+b)

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All Calculus AB Resources

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

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All Calculus AB Resources

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

Flashcards: Connect Position, Velocity, and Acceleration

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DMCA Complaint

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