Equations of Uniform Accelerated Motion

Equations of Uniform Accelerated Motion + Physics Mrs. Coyle / Mr. Davis

Distance / Displacement d = vt df = di + vt

Problem 1 Jared starts running at the 50 yard line. He runs at 9 yards per second for 4.5 seconds. Where does he end up? Does he make a touchdown?

Average Velocity vave = ½ (vf +vi): this is just how you take an average

Problem 2 Anita starts walking at 1mph and speeds up to 3mph by the time she is done over a two hour span. What is her average velocity? (Note that we can only answer this if she has a constant acceleration.)

Average Velocity vave = ½ (vf +vi): this is just how you take an average (This assumes constant acceleration: v = at) vf = vi + at

Displacement in terms of Average Velocity and Time d= vave t d= ½ (vf + vi) t

How do we derive d= ½ (vf + vi)t from the graph? vf Velocity (m/s) vi o Time (s) t • Hint: Area Under the Line=Displacement Δd or simply d

Average Velocity vave = ½ (vf +vi): this is just how you take an average v = at (This assumes constant acceleration) vf = vi + at

Displacement (d) in terms of vi , a, t d= ½ (vf + vi) t

Problem 3 Anita starts walking at 1mph and speeds up to 3mph by the time she is done over a two hour span. How far does she go? d= ½ (vf+ vi) t d= vave t

Problem 4 Anita starts walking at 1mph and speeds up to 3mph by the time she is done over a two hour span. How fast does she accelerate? vf = vi + at

Displacement (d) in terms of vi , a, t d= ½ (vf + vi) t, and vf = vi + at

Displacement (d) in terms of vi , a, t d= ½ (vf + vi) t, and vf = vi + at, so d= ½ ((vi + at)+ vi)t

Displacement (d) in terms of vi , a, t d= ½ (vf + vi) t, and vf = vi + at, so d= ½ ((vi + at)+ vi)t thus d = ½ (2vi + at)t

Displacement (d) in terms of vi , a, t d= ½ (vf + vi) t, and vf = vi + at, so d= ½ ((vi + at)+ vi)t thus d = ½ (2vi + at)t implying d= vit + ½ at2

Problem 5 Anita starts walking at 1mph and speeds up to 3mph by the time she is done over a two hour span. How far does she go? d= vit + ½ at2

Displacement (d) in terms of vi , a, t d= vit + ½ at2 for vi = 0

Final Velocity in terms of vi, a, d Again, starting with d= ½ (vf + vi) t. How do we get rid of t?

Final Velocity in terms of vi, a, d Starting with d= ½ (vf + vi) t, and remembering vf = vi + at

Final Velocity in terms of vi, a, d Starting with d= ½ (vf + vi) t, and remembering vf = vi + at which can be written as t= (vf – vi) / a

Final Velocity in terms of vi, a, d Starting with d= ½ (vf + vi) t, and remembering vf = vi + at which can be written as t= (vf – vi) / a, then d= ½ (vf + vi) (vf – vi) / a

Final Velocity in terms of vi, a, d Again, starting with d= ½ (vf + vi) t, and remembering vf = vi + at which can be written as t= (vf – vi) /a, then d= ½ (vf + vi) (vf – vi) /a which equals 2ad = (vf + vi) (vf – vi)

Final Velocity in terms of vi, a, d Again, starting with d= ½ (vf + vi) t, and remembering vf = vi + at which can be written as t= (vf – vi) /a, then d= ½ (vf + vi) (vf – vi) /a which equals 2ad = (vf + vi) (vf – vi) or 2ad = vf 2 – vi2

Final Velocity in terms of vi, a, d Again, starting with d= ½ (vf + vi) t, and remembering vf = vi + at which can be written as t= (vf – vi) /a, then d= ½ (vf + vi) (vf – vi) /a which equals 2ad = (vf + vi) (vf – vi) or 2ad = vf 2 – vi2 or more commonly vf 2 = vi2 + 2ad

Problem 5 Anita is in training to be an Olympic class walker. She is slowly increasing her speed as she walks as has been discussed. At her rate of acceleration, how far will she have gone when her speed reaches 5mph? vf 2 = vi2 + 2ad

Final Velocity in terms of vi, a, d vf 2 = vi2 + 2ad for vi=0

Equations of Motion for Uniform Accelerated Motion vf= vi+ at vavg = ½ (vf +vi) d= ½ (vf + vi)t d= vit + ½ at2 vf2 = vi2 + 2ad • d is the displacement (or Δd) • Assume that ti=0 • a is assumed constant

Solving Kinematics Problems • Draw a labeled vector diagram showing the positive and negative direction. • Make a list of the givens (include signs as needed) and unkown. • Decide what equation(s) you should use. • Write the equation(s) and solve for the unknown. Always include units in your first substitution and in your final answer.

Problem 6 A rocket travelling at +95m/s is accelerated uniformly to +150m/s in 10s. What is the displacement? Answer:1,225.m

Problem 7 An airplane has a minimum take off velocity of 80m/s. How long should the runway be, if the airplane can accelerate on the ground at 3m/s2 ? Answer: 1,067m

Problem 8 An airplane landing at +100m/s, comes to a stop in 30s. • What is the acceleration? • How far did it travel on the runway before it stopped? Answer: -3.3m/s2, 1,500m (Note: don’t round the acceleration)

Equations of Uniform Accelerated Motion