Physics 2514Lecture 6

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

Physics 2514 – p. 1/19

Goals

The goal of this lecture is to apply the equations derived in theprevious lecture to a few situations.We will also introduce the concept of acceleration caused bygravity.

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Review

Basic equations:

vavg =∆x

∆taavg =

∆v

∆tv(t) =

ds(t)

dta(t) =

dv(t)

dt

Average VelocityInstantaneous Velocity

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Review

Equations for Constant Acceleration

s(t) =1

2at2 + v0t + s0 v(t) = at + v0

v2 = v20 + 2a(s − s0)

Position vs. Time

PSfrag replacements

0

0t

s(t

)

s0

s(t) = 1

2at2 + v0t + s0

Velocity vs. Time

PSfrag replacements

0

0t

v(t

)

v0

v(t) = at + v0

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Constant Acceleration

At times it is useful to have a relation between the position andvelocity without the time.

Start with the equations for velocity and time and solve forthe time

v = at + v0 ⇒ t =v − v0

a

Substitute t into the expression for the position

t =v − v0

a

x =1

2at2 + v0t + x0

⇒ v2 − v20 = 2a(x − x0)

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Clicker

The plot below shows the position of an object as a function oftime. Determine the time t when the object changes direction.Assume that x(0) = −10 m, v(0) = −15 m/s, and a = 5 m/s2 witha constant.

Position vs. Time

PSfrag replacements

0

0t

x(t

)

x0

x(t) = 1

2at2 + v0t + x0

1. t = 6 s2. t = (3 +

√13) s

3. 3 s4. 0 s5. −1 s

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Example

Two trains initially d = 7.0 km apart approach each other onparallel tracks. Train 1 has a constant speed of v = 75 km/h withrespect to the ground. Train 2 is initially at rest, but acceleratestoward train 1 with a constant acceleration of a = 10 km/hr2.How long will it be before they reach each other?

1 2

Trains 7.0 km apart, train 1 has a constant speed of 75 km/h,train 2 initially at rest with constant acceleration 10 km/hr2 howlong until they are at the same location?

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Example

Trains 7.0 km apart, train 1 has a constant speed of 75 km/h,train 2 initially at rest with constant acceleration 10 km/hr2 howlong until they are at the same location?

y

x

PSfrag replacements−avo

x0

t0 = 0 h

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Example

Trains 7.0 km apart, train 1 has a constant speed of 75 km/h,train 2 initially at rest with constant acceleration 10 km/hr2 howlong until they are at the same location?

KnownTrain 1

a1 = 0 km/h2

v1(0) = 75 km/h,x1(0) = 0 km

Train 2a2 = −10 km/h2

v2(0) = 0 km/h,x2(0) = 7 km

Unknownx and t when they meet

y

x

PSfrag replacements−avo

x0

t0 = 0 s

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Example

When do they meet?They meet when their positions are the same.

The equations:

x1(t) = v1,0t train 1

x2(t) =1

2a2t

2 + x2,0 train 2

v1,0 = 75 km/h, a2 = −10 km/h2, x2,0 = 7 kmy

x

PSfrag replacements

−avo

x0

t0 = 0 h

Physics 2514 – p. 10/19

Example

When do the meet?They meet when their positions are the same.The condition: x1(t) = x2(t)

v1,0t =1

2a2t

2 + x2,0 ⇒ 0 =1

2a2t

2 − v1,0t + x2,0

t =v1,0 ±

√

v21,0 − 2ax2,0

a⇒

{

0.0927 hr−15.093 hr

Only top + result makes sense for this problem.v1,0 = 75 km/h, a2 = −10 km/h2, x2,0 = 7 km

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Gravity

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Gravity

In the 17th century Galileo carried out a series of experimentsthat found that all objects independent of mass fall toward theEarth at the same rate.

The rate at which objects fall is g = 9.8 m/s2(toward theEarth). This assumes that we neglect resistance due to theEarth’s atmosphere, and that we are close to the Earth’ssurface.

The value of g is only correct near the surface of theEarth.The value of g is different for other astronomical objects(Earth’s moon g = 1.6 m/s2, for Mars g = 3.7 m/s2, forJupiter g ≈ 26 m/s2).

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Gravity

The acceleration toward the Earth is given by ay = −g withg = 9.8 m/s2

Where we have assumed that our coordinate system issuch that the y axis is perpendicular to the Earth’ssurface and points upward.

Mathematical Description

y(t) = −

1

2gt

2 + v0yt + y0

vy(t) = −gt + v0y

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Assignment

Finish reading Chapter 2 and read Chapter 3 for the next lectureFriday’s lecture will give a brief introduction (review) of vectors,and discuss motion on an incline plane.Discussion sections will meet tomorrow.

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