Suggested problems
15
The famous four formulae
v = v0 + at
(1)
1 2
x − x0 = v0t + at
2
(2)
1
x − x0 = (v + v0)t
2
(3)
v2 − v20 = 2a(x − x0)
(4)
You will definitely need to know these!
16
Example: In the 100 m dash a sprinter accelerates from rest to a top speed
with an acceleration of 2.68 m/s2. After achieving top speed, he runs the
rest of the race at constant speed. If the total race is run in 12.0 s, how far
does he run while accelerating?
Put T = duration of acceleration. Remaining time is 12 – T seconds.
Acceleration phase
Acceleration: a = 2.68 m/s2
Final speed: v = aT
(1)
Distance covered: x1 = aT2/2 (2)
Coasting phase
Remaining time = 12 – T = (100 – x1)/v = (100 – aT2/2)/(aT)
So,
aT(12 – T) = 100 – aT2/2
aT2/2 – 12aT + 100 = 0
Substitute for a, solve: 1.34T2 – 32.16T + 100 = 0
17
For a quadratic equation: ax2 + bx + c = 0
the solution is:
√
−b ± b2 − 4ac
x=
(Appendix C4)
2a
Solve: 1.34T2 – 32.16T + 100 = 0
√
32.16 ± 32.162 − 4 × 1.34 × 100
T=
seconds
2 × 1.34
32.16 ± 22.32
=
seconds
2.68
= 3.67 s, or 20.32 s (longer than the 12 s race!)
So, T = 3.67 s, x1 = aT2/2 = 18 m during acceleration
18
Free Fall - negligible air resistance
Which way is up? – two choices (equally good)
y
a = –g = –9.80 m/s2
Acceleration by gravity is downward, opposite
in direction to the y-axis.
a = g = +9.80 m/s2
y
Acceleration by gravity is downward, in same
direction as the y-axis.
19
Example: A coin is dropped from the top of a building 427 m high.
Ignoring air resistance, at what speed does the coin hit the ground, how
long to reach it?
y
yo = 427 m
a = –g = –9.80 m/s2 vo = 0
Speed just before hitting ground
y=0
Time to reach ground
v = v0 + at
−91.5 = 0 + (−9.8)t
t = 91.5/9.8 = 9.34 s
(4)
v2 = v20 + 2a(y − y0)
= 0 + 2(−9.8)(0 − 427)
v = −91.5 m/s (downward)
(1)
1 2
[Or, y − y0 = − gt ]
2
20
Example: A diver springs upward with an initial speed of 1.8 m/s from a
3 m diving board. Find the speed with which he strikes the water and the
highest point he reaches.
y
Highest point: y = h, v = 0
a = –g
Diving board: y = 0, v0 = +1.8 m/s
Water level: y = –3 m, v = vw
On entry into water, y = –3 m:
v2w = v20 + 2ay
= 1.82 + 2(−9.8)(−3)
vw = 7.88 m/s
At highest point, v = 0:
v2 = v20 + 2ay
0 = 1.82 + 2(−9.8)h
h = 1.82/(2 × 9.8) = 0.17 m
21
A tile falls from rest from the top of a building. The tile takes 0.2 s to fall
past a window that is 1.6 m high. From what height above the window
did the tile start to fall?
Top of building
A
v0 = 0, a = –g, y0 = h
y
B
v1, t = 0, y1 = 0
C
v2, t = 0.2 s, y2 = -1.6 m
Window
1
At C, y2 = y1 + v1t − gt 2
2
1
−1.6 = 0 + 0.2v1 − g × 0.22
2
At B, v21 = v20 + 2gh
v1 = −7.01 m/s
7.012 − 0
h=
= 2.51 m
2 × 9.8
22
A hot air balloon is ascending straight up at a constant speed of 7.0 m/s.
When the balloon is 12.0 m above the ground, a gun fires a pellet straight
up from ground level with an initial speed of 30.0 m/s. At what two places
are the balloon and pellet at the same height at the same time?
vb = 7 m/s, constant
yb = y + vbt
1 2
Pellet: y p = 0 + v1t − gt
2
y0 = 12 m
Pellet and balloon meet when yb = yp
y
v1 = 30 m/s
y=0
a = -g
1
y0 + vbt = v1t − gt 2
2
1 2
gt + (vb − v1)t + y0 = 0
2
or, 4.9t 2 − 23t + 12 = 0
√
23 ± 232 − 4 × 4.9 × 12
t=
= 0.598 s, or 4.096 s
9.8
23
t = 0.598 s, or 4.096 s
yb = y + vbt = 12 + 7t
So, yb = 16.19 m, or 40.67 m
24