Ch 2 – Motion In 1-D



2.3 Motion With Constant Acceleration

Based on what you’ve seen thus far, it appears that motion with constant acceleration is common.

If a particle moves with constant acceleration, a is defined as: [pic]

For particles that start at time1 = 0, and experience a change in velocity,

[pic]

Manipulation of the far left and far right of the equation gives us

[pic]

Q: What does this equation really mean?

A: Velocity is due to not only the initial velocity, but also to any acceleration the particle experiences over time.

[pic]

Whenever the acceleration is constant, (velocity varies linearly with time), the average velocity is then the average of the initial and final velocities for the time interval in question.

Constant velocity vs constant acceleration (2015): (6:54)

[pic]

The displacement can then be rewritten as:

∆x = x – x0 = vavet = ½ (v0 + v)t

If you substitute v = v0 + at, you get

∆x = x – x0 = vavet = ½ (v0 + v)t = ½ (v0 + v0 + at)t = v0t + ½ at2

Q: What relationship does the above formula have to the integral (anti-derivative) of v = v0 + at with respect to time?

A: They are identical!!

This means the displacement can be rewritten as:

[pic]

One last tweak:

If the initial position is (0, 0),

∆x = v0t + ½ at2

If the initial position is not (0, 0), then you have to add an x0 to the equation to give:

x = x0 + v0t + ½ at2

OK, just one morr derivation…

If ∆x = vavet and vave = ½ (v0 + v)t, and if you solve v = v0 + at for “t” to get [pic], then you would have:

∆x = vavet = ½ (v0 + v)t = ½ (v0 + v) [pic] = [pic]

Using the extreme left and extreme right, and solving for v2 gives:

[pic]

This section contains some of the most important base equations for kinematics!

Know them like the back of your hand!!! [pic]

See p 28, Ex 2-9. Remember to read the Remarks section! Study the correlating graphs!

Upon graduation, a joyful physics student throws her cap straight upward with an initial speed of 14.7 m/s. Given that the acceleration is 9.81 m/s2 downward (neglecting air resistance),

a) How long does it take to reach its highest point? (Estimate it conceptually first!!)

b) What is the distance to the highest point?

c) Assuming the cap is caught at the same height from which it was released, what is the cap’s total time in flight?

[pic] [pic]

If not yet seen, let’s see this video relating d, v, a, j and graphical analysis. (



Here’s a video on position, velocity and acceleration from Bozemon Science (2015 7:54)



Last year, we studied stopping distance as a function of KE.

See Ex 2-10, p 20

On a highway at night you see a stalled vehicle and brake your car to a stop with an acceleration of magnitude 5 m/s2. What is the car’s stopping distance if its initial speed is

a) 15 m/s

b) 30 m/s

c) What is the mathematical relationship between these values? Does that make sense?

Ex 2-10 p 29-30 Solved Graphically

[pic]

Either way, this gives you a right triangle with a height of 15 m/s and a length of 3 s.

Remember – the area under a velocity – time graph is the displacement. Since the shape (graphically) is a triangle, just use

A = ½ b(h

A = ½ b(h

= ½ (3 s)(15 m/s)

= 22.5 m

Note: The units work out, as they should!!! (

See Ex 2-12, p 31. Read the Remarks section!

[pic]

See Ex 2-14, p 32 – 33

h1 = hi,1 + vi,1t1 + ½ ai,1 t1 2 h2 = hi,2 + vi,2t2 + ½ ai,2 t2 2

0 0

h1 = hi,1 + vi,1t1 + ½ ai,1 t1 2 h2 = hi,2 + vi,2t2 + ½ ai,2 t2 2

Since h1 = h2, set the equations equal to each other. If the start time is 0.69 s, then 1.68 s later (t2) would be 2.37 s. Also, since they are launched and caught at the same height, vi,1 = vi,2. Can we then cancel them out, as well?

No – since they are being multiplied by different times, but they will be like terms!

vi,1t1 + ½ ai,1 t1 2 = vi,2t2 + ½ ai,2 t2 2

vi t1 + ½ ai,1 t1 2 = vi t2 + ½ ai,2 t2 2

Substitute the values, solve for vi, plug back in, and solve for h

vi (0.69 s) + ½ ( - 9.81 m/s2)(0.69 s)2 = vi (2.37 s) + ½ ( - 9.81 m/s2)(2.37 s)2

vi = 15.009 m/s

hf = vi t1 + ½ ai t1 2

= (15.009 m/s)(0.69) + ½ (- 9.81 m/s2)(0.69 s)2

= 8.02 m

See Ex 2-15, p 33

[pic]

xs = xs,i + vs,i t + ½ as,i t2 xp = xp,i + vp,i t + ½ ap,i t2

0 0 0 0

xs = xs,i + vs,i t + ½ as,i t2 xp = xp,i + vp,i t + ½ ap,i t2

xs = vs,i t xp = ½ ap,i t2

You can solve this graphically. To do so, call xs “y1” on your calculator, and call xp “y2” on your calculator. Time is now “x” on calculator.

Enter each equation, do 2nd, calc, intersect. Select curve 1 and curve 2. The 2nd intersection point will give you your time!

y1 = 25 x

y2 = ½ (5) x2

The intersection occurs at (10, 250) ( t = 10 sec

Then, substitute t = 10 sec into vp = vi, p + apt

0

vp = vi, p + apt

vp = (5 m/s2)(10 s) = 50 m/s

On an exam, you would have to tell exactly what you are doing (window, 2nd-calc-intersect, et cetera), and draw out your screen.

You can do the same thing with EX 2-16. Simply add the constant of 25 to y2 (

Or, you can solve it using the same kinematics equations with an algebraic focus.

EX 2-16 Done algebraically:

xs = xs,i + vs,i t + ½ as,i t2 xp = xp,i + vp,i t + ½ ap,i t2 + 25 m

0 0 0 0

xs = xs,i + vs,i t + ½ as,i t2 xp = xp,i + vp,i t + ½ ap,i t2 + 25 m

xs = vs,i t xp = ½ ap,i t2 + 25 m

Since you added the 25 m to the police officer’s side, you can now set the equations equal to each other. Substitute in the values, and you get

25 m/s ∙ t = ½ (5 m/s2) ∙ t2 + 25 m Use QUADFOUR

0 = 2.5 m/s2 ∙ t2 – 25 m/s ∙ t + 25

t = 8.87 sec (and 1.13 sec – which you reject)

Scroll down to see EX 2-17and EX 2-18

See Ex 2-17, p 34-35

[pic]

EX 2-18, p 35 – 36

Consider EX 2-17. Assume the velocity of the elevator is 16 m/s up when the screw separates from the ceiling.

a. How far does the elevator rise while the screw is falling? How far does the screw fall?

b. What are the velocities of the screw and the elevator at impact?

c. What is the velocity of the screw relative to the floor at impact?

a. he = h0 + v0t + ½ ae t2

hf = h0 + v0t + ½ ae t2

= (16 m/s)(0.659 s) + ½ (4 m/s2)(0.659 s)2

= 11.4 m Since the elevator is 3 m from the floor, 11.4 m – 3 m = 8.4 m

= 8.4 m for the screw

b. vf,e = v0,e + ae t vf,s = v0,s + as t

= (16 m/s) + 4 m/s2 (0.659 s) = 16 m/s + (-9.81 m/s2) (0.659 s)

= 18.636 m/s = 9.53 m/s

c. + 9.53 m/s – 18.636 m/s = - 9.106 m/s

-----------------------

As you can see in the diagram to the right, the particle has an initial positive velocity, as given by its y-intercept. Because there is a positive slope to this linear graph, it obviously experiences a positive, constant acceleration.

Vel (m/s)

Time (s)

5

There are 3 ways to determine the time it will take to stop. They work because the

deceleration is constant.

Way 1: Conceptual

If the object starts at 15 m/s and decelerates at a rate of 5 m/s/s, then after 1 full sec, it will be going 10 m/s,

after 2 full secs, it will be going 5 m/s, and after 3 full secs, it will be going 0 m/s, (it takes 3 sec to stop.

Way 2: Unit analysis

15 m/s = 3 s to stop

5 m/s/s

Way 3: Algebraic / graphical

Start at v = + 15 m/s. go down 5, right 1

(a = - 5 m/s2) until you get to the x axis.

[?]'(-MŽ¼ËÎÏâãäåæç 6 7 J K L M ” • ¬ ­ õêáØÏØÆØƸة–¸ÆØÆ‹Æ}Æn[}Æ}Æ}$jM[?]hIuhIu5?CJEHèÿThe x-intercept = 3 s

In a crash test, a car traveling 100 km/h hits an immovable concrete wall. What is its acceleration?

A car is speeding at 25 m/s in a school zone. A police car starts from rest just as the speeder passes and accelerates at a constant rate of

5 m/s2.

a) When does the police car catch the speeding car?

b) How fast is the police car traveling when it catches up with the speeder?

Elevator Screw

h0 = 0 m h0 = 3 m

hf = hf hf = hf

v0 = 0 m/s v0 = 0 m/s

ae = 4.00 m/s2 as = - 9.81 m/s2

hf = h0 + v0t + ½ ae t2 hf = h0 + v0t + ½ as t2

0 0 3 0

hf = h0 + v0t + ½ ae t2 hf = h0 + v0t + ½ as t2

Since both are equal to hf, set them equal to each other

½ae t2 = 3 + ½ as t2

½ae t2 - ½ as t2 = 3

Factor out ½ t2 ½ t2 (ae – as) = 3

Divide both sides by ½ t2 (ae – as) = 3∙2

Divide both sides (ae – as) t2 (ae – as) = ___6___

(ae – as) (ae – as)

Substitute and take the t2 = ____6___ = __6__

square root of both sides (4 - - 9.81) 13.81

t = 0.659 s

While standing in an elevator, you see a screw fall from the ceiling. The ceiling is 3m above the floor. How long does it take for the screw to hit the floor if the elevator is moving up and gaining speed at a constant rate of af = 4.0 m/s2?

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