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.