Physics 151 Week 5 Day 1

Topics Area under a velocity graph Constant acceleration equations Pictorial (a.k.a Picture) diagram Solving Motion Problems using Strategic

Problem Solving (SPS) Approach

An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. Which graph best represents the vertical velocity of the arrow as a function of time? Ignore air resistance; the only force acting is gravity.

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Checking Understanding

An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. Which graph best represents the vertical velocity of the arrow as a function of time? Ignore air resistance; the only force acting is gravity.

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Answer

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Velocity to Position: Example 1

Find the position at times t = 1, 2, 3, 4, and 5 seconds.Assume x (t = 0 sec) = 0 m.

What is the displacement of the object between t = 1s and t = 3 s?

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Velocity to Position: Example 2

Describe in words the motion of the object whose velocity graph is given below. What is happening at t = 2 s?Draw a motion diagram of the objects motion.

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Velocity Graphs to Position Graphs: •Watch transitions between motion intervals

• If vx vs. t graph is continuous, slope of x vs t has no bumpsmust be smooth

• (at transition slope from left must = slope from right)

•Make sure slope of position graph matches velocity graph

Example

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Start your engine (velocity and acceleration)

On the on ramp to Interstate 25 a car accelerates at 7 m/s2 from rest.

What is the car's velocity at 1 second after it starts from rest?

2 seconds?

3 seconds?

4 seconds?

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General Motion Model Graph Relationships

Δx = area under vx vs. t graphΔvx = area under ax vs. t graph

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Constant Acceleration Equations

Δ vx = area = bh = Δ t axΔvx = ax Δ t

⇒ v1x = v0x + ax Δ t (equation 1)

Δx = area under vx vs. tΔx = Area1 (square) + Area2 (triangle)

Δx = bh1 +1

2bh2 = Δ t v0x +

1

2Δ t Δv1x

Δx = Δ t v0x +1

2Δ t aΔt( )

⇒ Δx = v0x Δ t +1

2ax Δ t

2 (equation 2)

Area 1

Area 2

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Constant Acceleration Equations (Equation 3)

⇒ v1x = v0x + ax Δ t (equation 1)

ax Δ t =v1x −v0x (equation1)

Δ t =v1x −v0x

ax

Δx = v0xv1x − v0xax

⎛

⎝⎜⎞

⎠⎟+1

2ax

v1x − v0xax

⎛

⎝⎜⎞

⎠⎟

2

(equation 2)

Δx =v0xv1xax

−v0x2

ax+1

2

v1x2 − 2v0xv1x + v0x

2

ax

⎛

⎝⎜⎞

⎠⎟

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Constant Acceleration Equations (Equation 3)

Δx =v0xv1xax

−v0x2

ax+1

2

v1x2 − 2v0xv1x + v0x

2

ax

⎛

⎝⎜⎞

⎠⎟

2axΔx=2v0xv1x −2v0x2 + v1x

2 −2v0xv1x + v0x2( )

2axΔx=2v0xv1x −2v0xv1x −2v0x2 + v0x

2 + v1x21x =v1x

21x −v0x

2

v1x21x −v0x

2 =2axΔx

⇒ v1x21x =v0x

2 + 2axΔx equation3

Unit Check => ???

Pictorial Diagrams

Pictorial Diagram Example 1

Pictorial Diagram Example 2

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The Sprinter

A sprinter accelerates at 2.5 m/s^2 until reaching his top speed of 15 m/s. He then continues to run at top speed.

How long does it take him to run the 100 m dash?

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Down and Up

A ball is released on the left side at a height of 1 m on a frictionless 30 degree slope, |a| = 5 m/s2. At the bottom, it turns smoothly onto a 60 degree slope going back up, |a| = 8.66 m/s2.

What maximum height does it reach on the right side?