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} = (6–3.5)/(0–3.6) v_{ave} =  0.69 m/s

(b) slope = Δy / Δx slope points (0s, 5.3m); (5.2s, 0m) v_{ }= Δx / Δt v_{ }= (5.3–0)/(0–5.2) 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.

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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 à 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} (6–1 s) + ½ (Dv / Dt) t^{2} 140 = v_{G} 5 sec + ½ Dv 1 sec 140 = 5.5v_{G} ß See note at bottom à 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})

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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 (can’t 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 