Unit 2 Class Notes

Honors Physics

The Kinematics Equations (1D Equations of Motion)

Day 6

Mixed Review (part II)

x = 2,224.5m ttotal = m28.72 sec

Clearly this is a free-fall problem

Remember the free-fall assumptions:V1 = 0, a = -9.8m/s2, down is negative

Solve the appropriate equations:

x 12 at

2 v1t

50 12 ( 9.8)t 2 0t

t 3.19sec

v22 v1

2 2ax

v22 02 2( 9.8)( 50)

v2 31.2 ms

What must be neglected??? AIR RESISTANCE!!!

Again, this is a free-fall problem

Remember the free-fall assumptions:V1 = 0, a = -9.8m/s2, down is negative

Solve the appropriate equations:

x 12 at

2 v1t

x0 3 12 ( 9.8)(3)2 0(3)

x0 3 44.1m

How far will it fall from 3-4 seconds?

x 12 at

2 v1t

x0 4 12 ( 9.8)(4)2 0(4)

x0 4 78.4m

d3 4 d0 4 d0 3 78.14 44.1 34.3m

During the 20th second?

d19 20 d0 20 d0 19

1960 1768.9 191.1m

Clearly this is a throw up problem

Remember the assumptions: V2 = 0 (@ top), a = -9.8m/s2, down is negative

Make sure to draw a picture and label the points appropriately.

x 12 at

2 v1t

50 12 ( 9.8)t 2 0t

t 3.19sec

What must be neglected??? AIR RESISTANCE!!! 1

2

3

4

Clearly this is a throw up problem

Remember the assumptions: V2 = 0 (@ top), a = -9.8m/s2, down is negative

Make sure to draw a picture and label the points appropriately.Choose two points, and work between these two points.

1

2

3

4

choose points 1 & 2choose points 1 & 2

choose points 1 & 4choose points 1 & 3

1

2

3

4

Points 1-2

v1 = 10 m/s

a = -9.8 m/s2

v2 = 0 m/s

02 102 2( 9.8)x

v22 v1

2 2ax

x 5.1m

v2 v1 at

0 10 ( 9.8)t

t 1.02sec

Points 1-4

a = -9.8 m/s2

x = 0 m

v1 = 10 m/s

x 12 at

2 v1t

0 12 ( 9.8)t 2 10t

t 2.04 sec

OR ….Simply double the time from point 1 to point 2.

1

2

3

4

v1 = 10 m/s

a = -9.8 m/s2

x = 5m

x 12 at

2 v1t

5 12 ( 9.8)t 2 10t

t 0.876sec, 1.165sec

Solve this quadratic equation for the times Graph y = -4.9x2 + 10x - 5

Points 1-3

0 12 ( 9.8)t 2 10t 5

Hit “2nd Trace”, “ZERO”

Left Bound, Right Bound, Guess ….It is 5 meters above the ground on the way UP and on the way DOWN.

Points 1-2

v1 40.92 fts

1

2

3

Remember your assumptions, draw a picture, pick your points.

x = x2 – x1 = 26 ft

v1 = ?

a = -32.2 ft/s2

x 1= 4 ft

x2 = 30 ft

v2 = 0

v22 v1

2 2ax

02 v12 2( 32.2)(26)

x = x2 – x1 = -4 ft

a = -32.2 ft/s2

x 1= 4 ft

x2 = 0 ft

v1 = 40.92 ft/s

x 12 at

2 v1t

4 12 ( 32.2)t 2 (40.92)t

16.1t 2 40.92t 4 0

t .094 sec, 2.64 sec

Points 1-3

This is a chase problem!!!

x2 12 at

2 v1t x1

x2N = x2O

Use the chase equation:

But it’s also a CHALLENGING chase problem. Why? Because the “new” car undergoes two different motions (speeding up and then coasting).

12 at

2 v1taccelerating

1 2 4 3 4 12 at

2 v1tcoasting

1 2 4 3 4 x1

N

12 at

2 v1t x1 O

Notice how each different motion needs to be accounted for when writing the equation.

Plug in what you know

12 at

2 v1taccelerating

1 2 4 3 4 12 at

2 v1tcoasting

1 2 4 3 4 x1

N

12 at

2 v1t x1 O

12 ataccel

2 0taccelaccelerating

1 2 4 4 3 4 4 12 (0)tcoast

2 21tcoastcoasting

1 2 4 4 4 3 4 4 4 0 12 (0)t 2 13t 0

v1 = 0

v2 = 21 m/s

t = 40 sec

v2 v1 atHow do we find “a”?

21 0 a(40)

a .525 ms2

How do we plug in for the times?

tO = t (the total time)

taccel = 40 sec (this was given)

tcoast = t – 40 (the total time minus the accel time)

Solve

12 (.525)(40)2 0(40) 1

2 (0)(t 40)2 21(t 40) 0 12 (0)t 2 13t 0

4200 0 21t 8400 0 13t 0

420 8t

t 52.5sec

13

21

v (m/s)

Time (sec)40 t

When the areas under the curves are equal, the new car has caught the older one.

Atriangle Arectangle Arectangle

12 (40)(21) (t 40)(21) 13t

t 52.5sec

THE GRAPHICAL APPROACH!!!

TONIGHTS HWNONE ENJOY!!!