motion in one dimension notes and example problems
DESCRIPTION
Motion in One Dimension Notes and Example Problems. D t = time in seconds D x= horizontal distance in meters D y = vertical distance in meters v i = the initial velocity in m/s v f = the final velocity in m/s a = acceleration in m/s 2. The Variables:. Acceleration of Gravity. - PowerPoint PPT PresentationTRANSCRIPT
Motion in One Dimension
Notes and Example Problems
t = time in secondsx= horizontal distance in metersy = vertical distance in metersvi = the initial velocity in m/s
vf = the final velocity in m/s
a = acceleration in m/s2
Acceleration of Gravity
• If an object is falling towards the earth surface then
you use g or ag = -9.81 m/s2 .*This means that an object speeds up 9.8 m/s
every second.*It is negative because objects fall downward.
• If the object is simply moving, use normal equations to solve for displacement and acceleration.
vav= x
t a = (vf -vi)
t y = vi t + ½ at2
y or x, depending on what direction the object is accelerating
x = ½(vi + vf )t
vf = vi + at
vf2 = vi
2 + 2ax
Problem Solving Tips
• Anything going down is • Object starting from rest • Object that is stopping • Object at its peak when thrown up • When dropping an object, we know
• Time up = time down in vertical displacement
Mickey drops Pluto off the kitchen table and, being the smart mouse that he is, he calculates that it takes him 0.53 seconds to hit the floor. How high is the table?
Problem Set-Up
• Given:
t = 0.53s not stated, but still there vi = 0 m/s
a = -9.8 m/s2
• Unknown:
y• Equation:
y = vi t + 1/2 at2
Solution
vi= 0 m/s so
y = vi t + 1/2 at2
since vi = 0 y = 1/2 at2
y = 1/2 (-9.8 m/s2)(0.53s)2
y = -1.376 m (falling down)• So the table is approximately 1.4 meters
high.
Tom needs to get away from a bomb that Jerry has set. If he uniformly accelerates
from rest at a rate of 1.5 m/s2, what will his final velocity be when he gets 25 meters
away?
Tom and Jerry Problem Set-Up
• Given:
vi = 0 m/s
x= 25 m
a = 1.5 m/s 2 • Unknown:
vf
• Equation:
vf2 = vi
2 + 2ax
Tom and Jerry Problem Solution
vf2 = vi
2 + 2ax
vf2 = (0 m/s)2 + 2(1.5 m/s2)(25 m)
vf2 = 75 m2/s2
vf = 8.7 m/s
Tom will be traveling at 8.7 m/s.
Swiper, thinking he is clever as usual, decides to get Dora once and for all. He gets on a one-person rocket and reaches a constant velocity of 16 m/s. Suddenly he realizes he is going to hit a brick wall (63 meters away) which he is unable to turn away from. He shuts the rocket off and coasts to a stop. Calculate the acceleration he will have to undergo in order to stop just before the wall.
Problem Set-Up
• Given:
vi = 16 m/s vf = 0 m/s
x = 63 m • Unknown:
a • Equation:
vf2 = vi
2 + 2a x
Solution
vf2 = vi
2 + 2a x
(0 m/s)2 = (16 m/s)2 + 2 a (63 m) - 256 m2/s2 = 2 a (63 m)
2 a = - 256 m2/s2
63 m
2 a = -4.06 m/s2
a = -2.03 m/s2
a = -2.0 m/s2
Wile E. Coyote pushes a piano off a cliff attempting to hitthe Road Runner. If it takes 3.5 seconds for the piano tohit the ground, calculate a) the speed at which the piano istraveling just before it hits the ground and b) the height ofthe cliff.
Road Runner Problem Set-Up
• Given:
vi = 0 m/s a= -9.8 m/s2
t = 3.5 s• Unknown:
vf and y
• Equation:
a = (vf - vi)/ t
Road Runner Problem Solution
PART a)
a = (vf -vi)/t but vi = 0 m/s so
(-9.8 m/s2) = (vf - 0 m/s)/ 3.5 s
(-9.8 m/s2) = vf / 3.5 s
= -34.3 m/s
The piano will hit the ground at a speed of -34 m/s, neglecting the affects of air resistance of course.
Road Runner Problem SolutionPART b)
• Unknown:
y
• Equation:
y = vi t + 1/2 at2
Road Runner Problem Solution
PART b)
y = vi t + 1/2 at2 but vi = 0 m/s so
y = 1/2 a t2
= 1/2 (-9.8 m/s2)(3.5 s) 2
= -60.025 m
The piano fell 60. meters.