Velocity MS-3 User Manual download pdf

math 131, applications motion: velocity and net change 1

Motion: Velocity and Net Change

In Calculus I you interpreted the ﬁrst and second derivatives as velocity and accel-

eration in the context of motion. So let’s apply the initial value problem results to

motion problems. Recall that

• s(t) = position at time t.

• s

(t) = v(t) = velocity at time t.

• s

(t) = v

(t) = a(t) = acceleration at time t.

Therefore

•

a(t) dt = v(t) + c

= velocity.

•

v(t) d t = s(t) + c

= position at time t.

We will need to use additional information to evaluate the constants c

and c

EXAMPLE 6.1. Suppose that the acceleration of an object is given by a(t) = 2 −cos t for

t ≥ 0 with

• v(0) = 1, this is also denoted v

• s(0) = 3, this is also denoted s

Find s(t).

SOLUTION. First ﬁnd v(t) which is the antiderivative of a(t).

v(t) =

a(t) dt =

2 −cos t dt = 2t −sin t + c

Now use the initial value for v(t) to solve for c

v(0) = 0 −0 + c

= 1 ⇒ c

= 1.

Therefore, v(t) = 2t −sin t + 1. Now solve for s(t ) by taking the antiderivative of v(t).

s(t) =

v(t) dt =

2t −sin t + 1 dt = t

+ cos t + t + c

Now use the initial value of s to solve for c

s(0) = 0 + cos 0 + c

= 3 ⇒ 1 + c

= 3 ⇒ c

= 2.

So s(t) = t

+ cos t + 2t + 2.

EXAMPLE 6.2. If acceleration is given by a(t) = 10 + 3t − 3t

, ﬁnd the exact position

function if s(0) = 1 and s(2) = 11.

SOLUTION. First

v(t) =

a(t) dt =

10 + 3t −3t

dt = 10t +

−t

+ c.

Now

s(t) =

10t +

−t

+ c dt = 5t

−

+ ct + d.

But s(0) = 0 + 0 −0 + 0 + d = 1 so d = 1. Then s(2) = 20 + 4 −4 + 2c + 1 = 11 so

2c = −10 ⇒ c = −5. Thus, s(t) = 5t

−

−5t + 1.

EXAMPLE 6.3. If acceleration is given by a(t) = sin t + cos t, ﬁnd the position function

if s(0) = 1 and s(2π) = −1.

SOLUTION. First

v(t) =

a(t) dt =

sin t + cos t dt = −cos t + sin t + c.

Now

s(t) =

−cos t + sin t + c dt = −sin t −cos t + ct + d.

But s(0) = 0 −1 + 0 + 0 + d = 1 so d = 2. Then s(2π) = 0 −1 + 2cπ + 2 = −1 so

2πc = −2 ⇒ c = −

+ 2. Thus, s(t) = cos t + sin t −

1 2 3 4 5 6

SOLUTION. Notice that v 5

Summary of Contents

Page 1

math 131, applications motion: velocity and net change 1Motion: Velocity and Net ChangeIn Calculus I you interpreted the ﬁrst and second derivatives a

Page 2

math 131, applications motion: velocity and net change 2Displacement vs Distance TravelledDEFINITION 6.1. The displacement of an object between times

Page 3

math 131, applications motion: velocity and net change 3The number line to the right shows that t3−5t2+ 4t ≤ 0 only [1, 4]. We can nowﬁnd the distance

Page 4

math 131, applications motion: velocity and net change 4The max height occurs when the velocity is 0 (when the ball stops rising):v(t) = −32t + 96 = 0

Page 5 - SOLUTION. Notice that v

math 131, applications motion: velocity and net change 5SOLUTION. We have: constant acceleration = a m/s2; v0= 0 m/s; s0= 0 m. Sov(t) = at + v0= atand

Page 6

math 131, applications motion: velocity and net change 6EXAMPLE 6.12. One car intends to pass another on a back road. What constant accel-eration is r

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Velocity MS-3 User Manual

Browse online or download User Manual for Fridge-freezers Velocity MS-3. Math 131, ApplicationsMotion: Velocity and Net Change

Summary of Contents

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