Chapter 2: Motion in One Dimension EXAMPLES. Example 2.1 Displacement. x 1 = 30 m x 2 = 10 m Displacement is a VECTOR. Example 2.2 Average Velocity & Speed. Suppose the person walks during 50 seconds. Displacement Distance (d) = 100m Average velocity: Average Speed: . X i.
Distance (d) = 100m
Instantaneous = average
Drive 65 miles/h on
Start your car!!!
Vyf(B) = vyi(A) + ayt(B)
0 = 20m/s + (–9.8m/s2)t(B)
t = t (B) = 20/9.8 s = 2.04 s
ymax = y(B) = y(A) +vyi(A)t + ½ayt2
y(B) = 0 + (20m/s)(2.04s) + ½(–9.8m/s2)(2.04s)2
y(B) = 20.4 m
y(C) = y(A)+ vyi(A) t – ½ayt2
(Solving for t): t(20 – 4.9t) = 0
vyf(C) = vyi(A) + ayt (C)
Using position (C) as the reference
point the tat (D) position is not 5.00s.
It will be:
t (D) = 5.00 s – 4.08 s = 0.96
vyf(D) = – 29.0 m/s
y(D) = y(C) +vyi(C)t + ½ayt2
After Ball 1 reaches maximum height it falls back downward passing the student with velocity –vi . This velocity is the same as Ball 2 initial velocity, so after they fall through equal height h, their impact speeds will also be the same!!!