Physics General I
Durig Lewis
Example
A car is traveling at a speed of 60 mph when the driver applies the brakes and decelerates at a rate of 10 mph/s. How long does it take the car to come to rest?
Solution
a = -10 mph/s vi = 60 mph vf = 0 mph x - axis
v = vf vi = 0 60 = -60 mph
v a t
10 mph / s 60 mph t
10 mph / s t 60 mph t 60 mph 10 mph / s
t 6 s
Example
A car is traveling at a speed of 60 mph when the driver applies the brakes and decelerates at a rate of 10 mph/s. How far does the car travel before it comes to rest?
Example
x - axis Time
Distance Traveled = Speed Time Taken = Area under Graph
Speed
Example
Speed
Distance Traveled = Area under Graph = HeightBase = 60 mph 6 s
1 60 miles 6 s 2 hour 1 60 1609 m = 6 s 2 3600 s =
= 80.45 m = 80.45 3.281 ft = 264 ft 1 mile = 1609 m 1 m = 3.281 ft
Time
1 hour = 60 min. = 3600 s
The Equations of Motion
What do we care about knowing? Position (initial = x0, final = x) Velocity (initial = v0, final = v) Acceleration (a) Time (initial = 0, final = t)
The Equations of Motion
Change in Velocity Average Acceleration= Time Taken v a= t v - v0 a= t at = v - v0 v = v0 +at
The Equations of Motion
Speed
Time Distance Traveled = Area Under Graph
The Equations of Motion
Speed v - v0 v0 Time t
t(v - v0 )
1 2
= 1 tat 2 = 1 at2 2 = tv0
Total Area= v0t+ 1 at2 2 Change in Position= v0t+ 1 at2 2 x - x0 = v0t+ 1 at2 2
The Equations of Motion
Speed
v0 t
v Time
Area of Trapezoid= 1 (v0 +v)t 2 Change in Position= 1 (v0 +v)t 2 x - x0 = 1 (v0 +v)t 2
The Equations of Motion
v = v0 +at x - x0 = v0t+ 1 at2 2 x - x0 = 1 (v0 +v)t 2
x - x0 = 1 (v0 +v)t 2 (v - v0 ) 1 x - x0 = 2 (v0 +v) a 2a(x - x0 )=(v0 +v)(v - v0 ) 2a(x - x0 )= v0 v - v2 +v2 - vv0 0 2a(x - x0 )=-v2 +v2 0 v2 = v2 +2a(x - x0 ) 0
The Equations of Motion
v = v0 +at x - x0 = v0t+ 1 at2 2 x - x0 = 1 (v0 +v)t 2 v2 = v2 +2a(x - x0 ) 0
Example
Convert a speed of 60 mph to S.I. units.
(1 mile = 1609 m)
60 miles 60 1609 m speed= = =26.8 m/s hour 3600 s
Example
A person, driving a car, accelerates from rest to a speed of 25 m/s in 5 seconds. What is the acceleration?
x0 =0 m x =Don't Care v0 =0 m/s v =25 m/s a=? t =5 s v = v0 +at x - x0 = v0t+ 1 at2 2 x - x0 = 1 (v0 +v)t 2 v2 = v2 +2a(x - x0 ) 0
Example
A person, driving a car, accelerates from rest to a speed of 25 m/s in 5 seconds. What is the acceleration?
x0 =0 m x =Don't Care v0 =0 m/s v =25 m/s a=? t =5 s v = v0 +at 25= 0+a 5 25= a 5 25 =a 5 a=5 m/s2
Example
A person driving a car at a speed of 25 m/s brakes and comes to rest in 5 seconds. What is the acceleration?
x0 =0 m x =Don't Care v0 =25 m/s v = 0 m/s a=? t =5 s v = v0 +at 0 =25+a 5 -25= a 5 -25 =a 5 a=-5 m/s2
Example
A person, driving a car, accelerates from rest to a speed of 25 m/s in 5 seconds. How far does the car travel in this time?
x0 =0 m x =? v0 =0 m/s v =25 m/s a=Don't Care t =5 s v = v0 +at x - x0 = v0t+ 1 at2 2 x - x0 = 1 (v0 +v)t 2 v2 = v2 +2a(x - x0 ) 0
Example
A person, driving a car, from accelerates rest to a speed of 25 m/s in 5 seconds. How far does the car travel in this time?
x0 =0 m x =? v0 =0 m/s v =25 m/s a=Don't Care t =5 s x - x0 = 1 (v0 +v)t 2 x -0 = 1 (0+25)5 2 x = 1 (25)5 2 x = 62.5 m
What is this to two significant figures?
Survey
A boy riding a skateboard throws a ball vertically upward. Which of the following best represents the path of the ball? B A C
Stopping Distance
What is the stopping distance of a car traveling at 60 mph?
General Physics I
Vectors
Geometry
A a b B
AB = ab
Aa = Bb
Geometry
Given that the two triangles shown are similar, find the length of side A in the larger triangle.
A 2 cm 3 cm 5 cm
Geometry
2 cm 3 cm A
A 5 cm = 2 cm 3 cm 5 cm A= 2 cm 3 cm A =3.3 cm
5 cm
Geometry
A 2 cm 3 cm 5 cm
Trigonometry
a b
h A
H
B
aA = hH bB = hH aA = bB
Trigonometry
opposite sin= hypotenuse
hypotenuse adjacent
opposite
adjacent cos= hypotenuse opposite tan= adjacent
Trigonometry
Find the length of side Q in the following right-angled triangle: Q 25 opposite sin= hypotenuse sin25= 1.5 Q Qsin25=1.5 Q= 1.5 sin25 Q =3.55 cm
1.5 cm
Trigonometry
Problem Find the length of side R in the following right-angled triangle:
cos = adjacent hypotenuse
R
50
8.0 cm
R cos50= 8.0 8.0cos50=R R =5.14 cm
Vectors
y V x Vx Vy Vx cos= V
Vx = Vcos
sin=
Vy V
Vy = Vsin
Vectors
A force of 20 Newtons acts on an object at an angle of 25 above the x-axis. Find the x and y-components of the force. y
Fx =20cos25 Fx =18.1N Fy =20sin25 Fy = 8.45 N Fx 20 N 25 x Fy
Vectors
y
Notation
r F =(18.1, 8.45)N r F =18.1x+8.45y r F =18.1i+8.45 j
20 N 25
Fy
x Fx
Adding Vectors
Add the vectors A=(3,5) N and B = (2,7) N
rr A+B =(3, 5)+(2, 7) rr A+B =(3+2, 5+7) rr A +B =(5,12)N
Adding Vectors
y
r F1
r F2
r F1 =(F1cos1 , F1sin1 ) x r F2 =(F2cos2 , F2sin2 ) rr F1 +F2 =(F1cos1 +F2cos2 , F1sin1 +F2sin2 )
Question
Given that the two triangles shown are similar, what is the length of side P? a) b) c) d) e) 3.3 cm 7.5 cm 8.4 cm 10.4 cm None of the above
P 8 cm 5 cm
12 cm
Answer
Given that the two triangles shown are similar, what is the length of side P?
P
P8 = 5 12 8 P = 5 12
P =3.3 cm
8 cm
5 cm
12 cm
Question
In the triangle shown, what is the length of side A? a)2.1 cm b)2.3 cm c)4.5 cm d)10.7 cm e)None of the above
A 25 5 cm
Answer
In the triangle shown, what is the length of side A?
A tan25= 5
A 25 5 cm
5 tan25= A
A =2.3 cm
Question
What are the (x,y) components of the force vector shown? a)(30, 30) N b)(39, 46) N c)(46, 39) N d)(42, 42) N e)None of the above y 60 N
40 x
Answer
What are the (x,y) components of the force vector shown?
Fx = 60cos40 Fx = 45.96 N Fy = 60sin40 Fy =38.57 N r F =(46, 39) N
y
60 N
40 Fx
Fy x
Question
What is the angle in the triangle below (not drawn to scale)? a)3 b)30 c)35 d)55 e)None of the above 7 cm
4 cm
Answer
What is the angle in the triangle below (not drawn to scale)?
opposite sin = hypotenuse
sin = 4 7
7 cm
4 = sin 7
-1
=35
4 cm