Lecture-3 (Dr. M Fadhali- General Physics)

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Dr. M. Fadhali

Motion with Constant Acceleration

The average velocity of an object during a time interval t is

The acceleration, assumed constant, is

Dr. M. Fadhali

Motion at Constant AccelerationIn addition, as the velocity is

increasing at a constant rate, we know that

Combining these last three equations, we find:

We can also combine these equations so as to eliminate t:

We now have all the equations we need to solve constant-accelerationproblems.

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• Recall that Since a is constant, we can integrate this using the above rule to find:

• Similarly, since we can integrate again to get:

1-D Motion with constant acceleration

v =dxdt

v = a dt = a dt = at + v0∫∫

x = v dt = (at + v0 )dt =1

2∫∫ at 2 + v0t + x0

Dr. Mohamed Al- Fadhali

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Useful Formula

Eliminating t:

x = x0 + v0t +1

2at 2l Solving for t:

x = x0 + v0

v − v0

a

+1

2a

v − v0

a

2

v2 − v02 = 2a(x − x0)

• Simplifying:

Dr. Mohamed Al- Fadhali

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In another way

a =dv

dt=

dv

dx⋅dx

dt

v2 −v02 = 2a(x− x0)

(chain rule)

a = v ⋅dv

dxa ⋅dx = v⋅dv

a d x = x 0

x

∫ a d xx 0

x

∫ = v ⋅ d vv 0

v

∫

⇒

⇒ a (x -x 0 ) =12

(v 2 − v 02 )

or

a = constant( )

Dr. Mohamed Al- Fadhali

Dr. M. Fadhali

Ex. 1 At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s2. At this rate,

how long does it take to accelerate from 80 km/h to 110 km/h?

.

The time can be found from the average acceleration, va

t∆

=∆

( )2 2

1m s30 km h

3.6 km h110 km h 80 km h5.208s 5 s

1.6 m s 1.6 m sv

ta

∆ −∆ = = = = ≈

Ex.2 A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach that speed?

The sprinter starts from rest. The average acceleration is found from

( )( )

( )( )

22 22 2 2 20

0 00

11.5 m s 02 4.408 m s 4.41m s

2 2 15.0 mv v

v v a x x ax x

−−= + − → = = = ≈

−the elapsed time is found by solving

00 2

11.5 m s 0 2.61 s

4.408 m sv v

v v at ta− −

= + → = = =

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Dr. M. Fadhali

Ex.3 A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 s. How far did it travel in that time?

The words “slowing down uniformly” implies that the car has a constant acceleration. The distance of travel is found form

( )00

21.0m s 0m s6.00 sec 63.0 m

2 2v v

x x t+ +

− = = =

2

20

0

0

0

0

vvtvxx

vvvbut

tvxxtvxx

+==−

+=

=−

+=

Galileo Free Falling

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Dr. M. Fadhali

Free Falling of Objects

Near the surface of the Earth,

all objects experience approximately the same

acceleration due to gravity.

This is one of the most common examples of motion with constant acceleration.

Dr. M. Fadhali

Falling Objects

In the absence of air resistance, all objects fall with the same acceleration, although this may be hard to tell by testing in an environment where there is air resistance.

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1-D Free-FallThis is a nice example of constant acceleration (gravity):In this case, acceleration is caused by the force of gravity:

– Usually pick y-axis “upward”– Acceleration of gravity is “down”:

yay = − g

v y = v0 y - g t

y = y0 + v0y t −1

2g t 2

vt

a

t

a y = − gy

t

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Problem:• The pilot of a hovering helicopter drops

a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance)

1000 mNeed to know: t and v(t)

Given: y(0) = 1000 m v(0) = 0

Acceleration?? a = - g = -10 m/s2

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First choose coordinate system.

Origin and y-direction.

Next write down position equation:

Realize that v0y = 0.

y = y 0 + v0y t1

2gt 2

y = y0 −12

gt 2

1000 m

y = 0

y

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Solve for time t when y(t)=0 given that y0 = 1000 m.

Recall

Solve for vy

y0 = 1000 m

y

T =2y0

g=

2 × 1000m

9.81m s 2 = 14.3s

y = y0 - 1

2gt 2

y = 0

vy2 - v0 y

2 = 2a(y - y0 )

vy = ± 2gy0

= −140 m / s

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Dr. M. Fadhali

Ex. 4 A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) How high does it go? (b) How long is it in the air?

(a) The displacement can be found from Eq. 2-11c, with x replaced by y .

( ) ( )( )

22 22 2 0

0 0 0 2

0 22m s2 0 25 m

2 2 9.80m sv v

v v a y y y ya

−−= + − → = + = + =

−

(b) The time of flight can be found from Eq. 2-11b, with x replaced by y , using a (c) displacement of 0 for the displacement of the ball returning to the height from which it was hit.

( ) ( )2 01 10 0 02 2 2

2 22m s20 0 0 , 4.5 s

9.80m sv

y y v t at t v at t ta

= + + = → + = → = = = =− −

The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of t = 4.5 s is the time to return to the original displacement. Thus the answer is t = 4.5 seconds.

Choose upward to be the positive direction, and take to be at the height where

the ball was hit. For the upward path,

0 0y =

0 22msv = 0v= at the top of the path, and 29.80m sa = −

Example : Runway DesignYou’re designing an airport. A plane that will use this airport must reach a speed of

vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If the runway isx = 150 m long, can this plane reach the speed of before it runs off the end of the

runway? (b) If not, what is the minimum length required for the runway?

Table of Knowns & UnknownsSolutions(a) Use Eq. (3):

v2 = (v0)2 + 2a(x – x0) v2 = 0 + 2(2.0)(150 – 0) = 600 m/s2

So v = (600)½ = 24.5 m/sNote that this ↑ means take the

square root! That matters! (b) Use Eq. (3) again with v = vmin = 27.8 m/s. Solve for x – x0 = [v2 – (v0)2]/(2a)x = [(27.8)2 – 0]/[2(2.0)]

So x = 193 m. To be safe, make the runway

200 m long!

(1)(2) (3)

(4)

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You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s if the car hits a brick wall. Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?

Air Bags

Known: x0 = 0 & v0 = 28 m/s & v = 0Car obviously stops when crash ends! ↑

Wanted unknown: t.But we don’t know acceleration a or distance x either! Estimate x = 1.0 m

This has to be a 2 step problem! First, useEq(3) to solve for a: 0 = (v0)2 + 2a(x – 0) so a = - (v0)2∕(2x) = - (28)2 ∕(2) = - 390 m/s2

This is a HUGE acceleration!! Now, use Eq (1) to solve for t: 0 = v0 + at so

t = - (v0) ∕a = 0.07 s !!!

(4)

Example: A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m.

(a) What can you say about the motion of the rocket after its engines stop?

(b) What is the maximum height reached by the rocket? (c) How long after lift-off does the rocket reach its maximum height?(d) How long is the rocket in the air?

(a) The rocket will continue upward, but start to decelerate due to the Earth’s gravitational field until the upward velocity reaches zero. The rocket then begins to fall back to the ground with an acceleration equal to the Earth’s surface gravity (i.e., 9.80 m/s2)

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Step 1:— First, we will need to calculate the velocity of the rocket whenthe engine is turned off: v1 ( we are given a1 = 2.00 m/s2, y1 = 150 m, & v0 = 50.0 m/s, so use

= (50.0 m/s)2 + 2(2.00 m/s2)(150 m)= 2.50 × 103 m2/s2 + 6.00 × 102 m2/s2

Now we need to calculate ymax ≅ y2.Step 2:—We will need y1, a2, and v1, also v2 = 0 (rocket comes to rest).Our initial velocity is now v1 and our final velocity is v2, so wecan write:

note that ymax = y2 and that the acceleration is now thedownward acceleration due to gravity: a2 = −g = −9.80 m/s2.

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c) How long does it take to reach ymax?Break the time into 3 parts:

t = t0 + t1 + t2

t = 0, which was our initial time, t1 is the time when the rocketreaches y1 and t2 is the time the rocket reaches ymax as measuredfrom y1, so use vf = vi + at:

where t(c) (the time it takes to fall from the maximum height tothe ground).

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January 7, 2011 Physics 114A - Lecture 4 23/19

Example: Speed of a Lava BombA volcano shoots out blobs of molten lava (lava bombs) from its summit.

A geologist observing the eruption uses a stopwatch to time the flight

of a particular lava bomb that is projected straight

upward. If the time for it to rise and fall back to

its launch height is 4.75 s, what is its initial

speed and how high did it go? ( g = 9.81 m/s2.)1 12

0 0 02 20 ( )x x v t gt x t v gt= + − ⇒ ∆ = = −1

0 2Either 0 or 0t v gt= − =1 1 2

0 2 2 (9.81 m/s )(4.75 s) 23.3 m/sv gt= = =2 2

0At maximum height, 0 2v v g x= = − ∆ 2 20

2

(23.3 m/s) 27.7 m2 2(9.81 m/s )vxg

∆ = = =