Ch 2.1
The
position of a pinewood derby car was observed at various times; the results are
summarized in the following table. Find the average velocity of the car for (a)
the first second, (b) the last 3 s, and (c) the entire period of observation.
t
(s) 0 
1.0 
2.0 
3.0 
4.0 
5.0 
x
(m) 0 
2.3 
9.2 
20.7 
36.8 
57.5 
(a) v_{ave} =
Δx
/ Δt v_{ave} = (2.30) / (10) v_{ave} = 2.3 m/s 
(b) v_{ave} =
Δx
/ Δt v_{ave} = (57.59.2) / (5 2) v_{ave} = 16.1 m/s 
(c) v_{ave} =
Δx
/ Δt v_{ave} = (57.50) / (5 0) v_{ave} = 11.5 m/s 
Ch 2.2 Instantaneous
velocity/speed Point A is at t
= 0 sec, B is at 1.7 sec, C is at 3.6 sec, D is at 4.7 sec, and E is at 5.4
sec. (a) Find the average velocity in the
time between A and C of the positiontime graph for a particle moving along
the x axis. (b) Determine the instantaneous velocity at point D by
measuring the slope of the tangent line shown in the graph. (c) At what point
the velocity zero? (d) Which point has
the greatest positive velocity? 


(a) a: (0 sec, 6 m) c: (3.6 sec, 3.5
m) v_{ave} =
Δx
/ Δt v_{ave} = (63.5)/(03.6) v_{ave} =  0.69 m/s 
(b) slope = Δy / Δx slope points (0s, 5.3m);
(5.2s, 0m) v_{ }= Δx / Δt v_{ }= (5.30)/(05.2) v =
1.02 m/s 
(c) Find where
tangent line is horizontal This occurs at Point B (d) Only Point A has a positive slope 





Ch
2.3 Acceleration
The left ventricle
of the heart accelerates the blood from rest to a velocity of 30 cm/s. The displacement of the blood during the
acceleration is +2 cm.
(a)
How
much time does blood take to reach its final velocity?
(b)
Determine
the acceleration (in cm/s^{2}.)
v_{ave} = Dx / Dt ½(30+0) = 2 / Dt Dt = 2/15 seconds 
a = Dv / Dt a = 30 / (2/15) a =
225 cm/s^{2} 
Ch 2.3 Acceleration
A particle moves along the xaxis according to the equation x
= 2 + 3t t^{2}, where x is in meters and t is in seconds. At t = 2 s, find (a) the position of the
particle, (b) its velocity, & (c) acceleration.
a) x = 2 + 3t t^{2} x = 2 + 3(3)
3^{2} x
= 2 m 
b) v = dx / dt v = 3 2t v =
3 m/s 
c) a = d^{2}x
/ dt^{2} a = dv/dt a =
2 m/s^{2} 
Ch 2.4 Motion
Diagrams
Draw a position versus time diagram for (a) an object moving
to the right at a constant speed; (b) and then the object moving to the right
and speeding up at a constant rate; (c) and then the object moving to the right
and slowing down at a constant rate; (d) and then the object moving to the left
and speeding up at a constant rate; and (e) lastly the same object moving to
the left and slowing down at a constant rate.

or 
Ch 2.5 1D Motion with Constant Acceleration
A particle moves along the xaxis. Its position is given by the equation x = t^{2}
 3t + 2 with SI units. Determine (a)
its position when it changes direction and (b) its velocity when it returns to
the position it had at t = 0.
(a) v = dx / dt v = 2t  3 From calculus class;
maximums occur when the derivative equals zero 
0 = 2t  3 t = 1.5 seconds x(t) = t^{2}  3t + 3 x = 0.75 m 
(b) x(0) = t^{2}  3t + 3 x(0) = 3 meters 3 = t^{2}  3t + 3 
t^{2} = 3t t = 3 sec v(3) = 2t  3 v(3) = 3 m/s 
Ch
2.6 Freely Falling Objects
If the sound of the splash, due to a rock dropped from rest
into a well, is heard 2.0 seconds later, how deep is the well? (v_{sound} = 340 m/s)
d_{r} =
½at_{r}^{2} + v_{o}t_{r}
+ d_{o} d_{r} = ½
10t_{r}^{2} d_{r} = 5
t_{r}^{2} 
d_{s} =
½at_{s}^{2} + v_{o}t_{s}
+ d_{o} d_{s} = v_{o}t_{s} 
5 t_{r}^{2} =
340(2  t_{r}) 5 t_{r}^{2} =
680  340t_{r} t_{r}^{2} + 68t_{r} = 136 (t_{r} + 34)^{2} = 136
+ (34)^{2} (t_{r} + 34)^{2} = 1292 t_{r}
= 34 ± 35.9444 t_{r}
= 1.9444 d_{r} = 5
t_{r}^{2} d_{r} = 18.9 meters 
We know that the distance
sound that travels is the same as the distance the rock travels. 
Given: v_{o} of
sound = 340 m/s t_{r} + t_{s} = 2 sec t_{s} = 2  t_{r} The rest is algebra 

d_{r} = d_{s} 5 t_{r}^{2} =
v_{o}t_{s} 5 t_{r}^{2} =
340t_{s} 
Ch 2.7 Kinematics
equations using calculus The time rate of change of acceleration is commonly known as
the jerk. Instead of acceleration
being constant as in our derivations, let acceleration be a variable and
jerk, J, be a constant. (a) Determine
1D expressions for acceleration a_{x}(t),
velocity v_{x}(t), and position x(t), given
that its initial acceleration, velocity, and position are a_{xi},
v_{xi}, x_{i}, respectively. 
_{} 
Ch 2 BONUS Two objects, A & B, are connected by a rigid rod that
has a length L. The objects slide
along perpendicular guide rails. If A
slides to the left with a constant speed, v_{A},
find the speed of B, v_{B}, when a = 80°. 


Given: A is at position +x, and
get smaller as B, at +y, gets larger,
thus: v_{a} = dx / dt +v_{b} = dy / dt need to take
derivative; x & y change as time
progresses 
x^{2} + y^{2} = L^{2} 2x dx/dt + 2y dy/dt = 0 2x (v_{a}) + 2y (+v_{b}) = 0 v_{b} = v_{a}
* x/y ΰ cot a =
x/y v_{b} = v_{a}
cot a v_{b} = v_{a}
cot 80° v_{b}
= 0.176 v_{a} 




Grader: for the above
problem, the students already had everything except the underline portions.
The underlined portions were left blank, so the students
only had to fill them in.
Ch 2 BONUS
In the African Savannah, a cheetah attempts to catch a
gazelle, but misses by less than 1 cm.
They both start running at this point.
The cheetah accelerates for 3 seconds; the gazelle accelerates for 1
second. It takes the cheetah 140 meters
and 6 seconds to catch back up to the gazelle. (a) What were their maximum
speeds? (b) What was the acceleration of each animal? (c) At 4 seconds, how far
ahead was the gazelle?
d_{G} = d_{o}
+ v_{o}t
+ ½ at^{2}: Gazelle 140 = v_{G} (61 s) + ½ (Dv / Dt) t^{2} 140 = v_{G} 5 sec + ½ Dv 1
sec 140 = 5.5v_{G} ί See note at bottom ΰ_{} v_{G} = 25.5 m/s a_{G} = Dv / Dt a_{G} =
25.5 / 1 a_{G} = 25.5 m/s^{2} 
d_{C} = d_{o}
+ v_{o}t
+ ½ at^{2}: Cheetah 140 = v_{C} (6 3
sec) + ½ (Dv / Dt) t^{2} 140 = v_{C} 3 sec + ½ Dv 3
sec 140 = 4.5v_{C} v_{C} = 31.1 m/s a_{C} = Dv / Dt a_{C} =
31.1 / 3 a_{C} = 10.4 m/s^{2} 
c) The gazelle is ahead by 11.4 m at 4 seconds 

d_{G} = ½
a_{G}*1^{2} + v_{G} (41 sec) d_{G} = 89.3 m 
d_{C} = ½
a_{C} 3^{2} + v_{C}
(4 3 sec) d_{C} = 77.9 m 
Note: At the end of the acceleration period is
the maximum velocity, v_{G} = Δv. The
reminder of the chase will be at this maximum speed. If the chase last for too long of a time,
the cheetah slows down appreciably. 
Ch 2.3 #16
An object moves along the xaxis according to the equation x(t) = 3t^{2} 2t + 3 meters. Determine (a) the average speed between t = 2
and 3 seconds; (b) the instantaneous speed at t = 2 and t = 3
seconds; (c) the average acceleration between t = 2 and 3 seconds; and (d) the
instantaneous acceleration at t = 2 and t = 3 seconds.
(a) x(t) = 3t^{2} 2t + 3 x(2) = 11 m; x(3)
= 24 m v_{ave} = Δx / Δt v_{ave} = (2411)/(32) v_{ave} = 13 m/s 
(b) v = dx / dt v = 6t 2 v(2) = 10 m/s v(3) = 16 m/s 
(c) a = Δv / Δt a = (1610)/(32) a = 6 m/s 
(d) a = dv / dt a = 6 a(2) = 6 m/s^{2} a(3) = 6 m/s^{2} 
Ch 2.5 #25
A particle moves along the xaxis. Its position is given by the equation x = 4t^{2}
+ 3t + 2 with SI units. Determine (a)
its position when it changes direction and (b) its velocity when it returns to
the position it had at t = 0.
v = dx / dt v = 8t + 3 From calculus class;
maximums occur when the derivative equals zero 
(a) v = 8t
+ 3 0 =
8t + 3 t = 3/8 seconds x(3/8) = 4t^{2} + 3t + 2 x = 2.56 seconds 
(b) x(0)
= 4t^{2} + 3t + 2 x(0)
= 2 meters 2 = 4t^{2} + 3t + 2 4t^{2} = 3t; t = 0.75 sec 
t = 0.75 seconds v(0.75) = 8t + 3 v(0.75) = 3 m/s 
Ch 2.7 #53
(Bonus)
Automotive engineers refer to the time rate of change of
acceleration as the jerk. If an object
moves in 1D such that its jerk J is constant, (a) determine expressions for
its acceleration a_{x}(t), velocity v_{x}(t), and position x(t), given that its initial
acceleration, velocity, and position are a_{xi},
v_{xi}, x_{i}, respectively. (b) Show that a_{x}^{2} = a_{xi}^{2}
+ 2J(v_{x} v_{xi})
J = Δa
/ Δt t = (a_{f}
a_{i}) /
J Since a is not constant a_{ave} =
Δv / Δt (a_{f} + a_{i})/2 =
(v_{f} v_{i}) / t (a_{f} + a_{i}) t
= 2(v_{f}
v_{i}) substitute in t from above (a_{f} + a_{i}) (a_{f}
a_{i}) / J = 2(v_{f}
v_{i}) (a_{f} + a_{i}) (a_{f}
a_{i}) =
2J(v_{f} v_{i}) a_{f}^{2} a_{i}^{2}
= 2J(v_{f} v_{i}) a_{f}^{2} = a_{i}^{2}
+ 2J(v_{f} v_{i}) 

Ch 2 #62 A test Rocket is fired vertically upward from a well. A catapult gives it an initial velocity of
80 m/s at ground level. Subsequently,
its engines fire and it accelerates upward at 4 m/s^{2} until it
reaches an altitude of 1000 m. At that
point its engines fail, and the rocket goes into free fall, with an acceleration
of 9.8 m/s^{2}. 
a) How long is the rocket in motion above the ground? (10 + 12 + 18.55) = 40.55 sec b) What is its maximum altitude? 1720
m c) What is its velocity just before it collides with the
Earth? 185.5
m/s 


A d = d_{o}
+ v_{o}t
+ ½ at^{2} 1000 = 0 + 80t +
½(4) t^{2} 2t^{2} +
80t 1000 = 0 t^{2} +
40t 500 = 0 (t  10 )
(t + 50 ) = 0 t = 10 seconds (cant equal 50 seconds) a_{y} =
Dv_{y} / Dt 4 = (v_{f} 80) / 10 v_{f} = 120 m/s 
B v_{i} of part B is v_{f}
of part A in part A, a = 4 m/s^{2},
in part B its in free fall,
a = 10 m/s^{2} a_{y} =
Dv_{y} / Dt 10 = (0
120) / Dt Dt = 12 s d = d_{o} + v_{o}t
+ ½ at^{2} d = 1000 + 120 (12) + ½(10) 12^{2} d = 1720 meters 

C Now we have a free fall problem starting from 1720 meters d = d_{o} +
v_{o}t + ½ at^{2} 0 = 1720 + 0 +
½ (10) t^{2} t = 18.55 s a = (v_{f} v_{i}) / t 10 = (v_{f} 0) / 18.55 sec v_{f} =
185.5 m/s 

Ch 2 #65
In a 100m race, Maggie and Judy cross the finish line in a
dead heat, both taking 10.2 seconds.
Accelerating uniformly, Maggie took 2 s & Judy 3 s to attain maximum
speed, which they maintained for the rest of the race.
a)
What was the acceleration of each sprinter?
b)
What were their respective maximum
speeds?
c)
Which sprinter was ahead at the 6 second
mark, and by how much?
d_{M} = d_{o} + v_{o}t + ½ at^{2}:
Maggie 100 = v_{M}
(10.2 2 sec) + ½ (Dv / Dt) t^{2} 100 = v_{M} 8.2
sec + ½ Dv t 100 = v_{M} 8.2
sec + ½ Dv 2 
d_{J} = d_{o} + v_{o}t + ½ at^{2}:
Judy 100 = v_{J}
(10.2 3 sec) + ½ (Dv / Dt) t^{2} 100 = v_{J} 7.2
sec + ½ Dv t 100 = v_{J} 7.2
sec + ½ Dv 3 

b) Thus v_{M} = Dv 100 = v_{M} 8.2
sec + ½ Dv 2 100 = v_{M} 8.2
sec + ½ v_{M} 2 100 = v_{M} 8.2 sec + v_{M}_{} 100 = v_{M} 9.2
sec v_{M} = 10.9 m/s 
a) a_{M} = Dv / Dt a_{M} = 10.9 / 2 a_{M} = 5.43 m/s^{2} 
b) Thus v_{M} = Dv 100 = v_{J} 7.2
sec + ½ Dv 3 100 = v_{J} 7.2
sec + ½ v_{J} 3 100 = v_{J} 7.2
sec + 3/2 v_{J} 100 = v_{J} 8.7
sec v_{J} = 11.5 m/s 
a) a_{J} = Dv / Dt a_{J} = 11.5 / 3 a_{J} = 3.83 m/s^{2} 
c) Maggie is ahead by 2.75 m at 6 seconds 

d_{M} = ½ a_{M}*2^{2} + v_{M}
(62 sec) d_{M} = 54.5 m 
d_{J} = ½ a_{J} 3^{2} + v_{J} (6 3 sec) d_{J} = 51.75 m 

Ch 2.1 #3
The position vs time for a certain
particle moving along the xaxis is shown in Figure P2.3 in your lecture
book. Find the average velocity in the
time intervals (a)
0 to 2 sec (b)
0 to 4 sec (c)
2 to 4 sec (d)
4 to
7 sec (e)
0 to 8 sec (a) v = Δx / Δt = (100) /(20) =
5 m/s (b) v = Δx / Δt = (50) / (40) =
1.25 m/s (c) v = Δx / Δt = (510) /(42) =
2.5 m/s (d) v = Δx / Δt = (55) /(74) =
3 1/3 m/s 

(e) v = Δx / Δt = (00) /
(80) = 0 m/s 
Ch 2.2 #9
Find the instantaneous velocity of the particle described in
Figure 2.3 at the following times:
a) t = 1 s b) t = 3 s c)
t = 4.5 s d) t = 7.5 s
By definition instantaneous
velocity is the slope of the point in question on the position vs time graph.
Another method
draw a normal at the point in
question. Then obtain the slope of a
line drawn perpendicular to the normal at the point in question.
In my solution, I am not use
either above technique; I am just obtaining the coordinates of two points on
the line that are very close to the point in question.
a)
t = 1 s v = Dx / Dt = (6 4)m / 0.4s = 5 m/s 
b)
t = 4.5 s v = Dx / Dt = (5 5)m / 0.1s = 0 m/s 
c)
t = 3 s v = Dx / Dt = (7 8)m / 0.4s = 2.5 m/s 
d)
t = 7.5 s v = Dx / Dt = (0 4)m / 0.8s = 5 m/s 
Ch 2.5 1D Motion with Constant Acceleration
A particle moves along the xaxis. Its position is given by the equation x = t^{2}
 3t + 2 with SI units. Determine (a)
its position when it changes direction and (b) its velocity when it returns to
the position it had at t = 0.
(a) v = dx / dt v = 2t  3 From calculus class;
maximums occur when the derivative equals zero 
0 = 2t  3 t = 1.5 seconds x(t) = t^{2}  3t + 3 x = 0.75 m 
(b) x(0) = t^{2}  3t + 3 x(0) = 3 meters 3 = t^{2}  3t + 3 
t^{2} = 3t t = 3 sec v(3) = 2t  3 v(3) = 3 m/s 
Ch 2.5 #27
A jet plane lands with a speed of 100 m/s & can accelerate
at a maximum rate of 5 m/s^{2} as it comes to rest.
a) From the instant the plane touches the runway, what is the
minimum time it needs before it can come to rest?
a_{ave} = Dv / Dt a_{ave} = (v_{f} v_{i}) / t 
5 = 0 100 / t t = 20 seconds 
b) Can this plane land at a small tropical island airport
where the runway is 0.8 km long?
v_{ave} = (v_{f}
+ v_{i}) / 2 
v_{ave} = (x_{f} x_{i}) / t;
where the distance, d = x_{f} x_{i} 
(v_{f}
+ v_{i}) / 2 = (x_{f} x_{i}) / t 50 m/s =
d /
20 sec d = 1000 m 
A minimum of 1000 meters is required for the jet
plane, thus an 800 meter runway is too short 