2. Motion in One Dimension Notes and Example Problems

3.  t = time in seconds  x= horizontal distance in meters  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

4. Acceleration of Gravity If an object is falling towards the earth surface then you use g or a g = -9.81 m/s 2. *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.

5.  v av =  x  t  a = (v f -v i )  t   y = v i  t + ½ a  t 2  y or  x, depending on what direction the object is accelerating   x = ½(v i + v f )  t  v f = v i + a  t  v f 2 = v i 2 + 2a  x

6. 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

8. Problem Set-Up Given:  t = 0.53s not stated, but still there v i = 0 m/s a = -9.8 m/s 2 Unknown:  y Equation:  y = v i  t + 1/2 a  t 2

9. Solution v i = 0 m/s so  y = v i  t + 1/2 a  t 2 since v i = 0  y = 1/2 a  t 2  y = 1/2 (-9.8 m/s 2 )(0.53s) 2  y = -1.376 m (falling down) So the table is approximately 1.4 meters high.

10. 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/s 2, what will his final velocity be when he gets 25 meters away?

11. Tom and Jerry Problem Set-Up Given: v i = 0 m/s  x= 25 m a = 1.5 m/s 2 Unknown: v f Equation: v f 2 = v i 2 + 2a  x

12. Tom and Jerry Problem Solution v f 2 = v i 2 + 2a  x v f 2 = (0 m/s) 2 + 2(1.5 m/s 2 )(25 m) v f 2 = 75 m 2 /s 2 v f = 8.7 m/s Tom will be traveling at 8.7 m/s.

14. Problem Set-Up Given: v i = 16 m/s v f = 0 m/s  x = 63 m Unknown: a Equation: v f 2 = v i 2 + 2a  x

15. Solution v f 2 = v i 2 + 2a  x (0 m/s) 2 = (16 m/s) 2 + 2 a (63 m) - 256 m 2 /s 2 = 2 a (63 m) 2 a = - 256 m 2 /s 2 63 m 2 a = -4.06 m/s 2 a = -2.03 m/s 2 a = -2.0 m/s 2

17. Road Runner Problem Set-Up Given: v i = 0 m/s a= -9.8 m/s 2  t = 3.5 s Unknown: v f and  y Equation: a = (v f - v i )/  t

18. Road Runner Problem Solution PART a) a = (v f -v i )/  t but v i = 0 m/s so (-9.8 m/s 2 ) = (v f - 0 m/s)/ 3.5 s (-9.8 m/s 2 ) = v f / 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.

19. Road Runner Problem Solution PART b) Unknown:  y Equation:  y = v i  t + 1/2 a  t 2

20. Road Runner Problem Solution PART b)  y = v i  t + 1/2 a  t 2 but v i = 0 m/s so  y = 1/2 a  t 2 = 1/2 (-9.8 m/s 2 )(3.5 s) 2 = -60.025 m The piano fell 60. meters.