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 / 5.0x (m) 0 / 2.3 / 9.2 / 20.7 / 36.8 / 57.5
(a)
vave = Δx / Δt
vave = (2.3-0) / (1-0)
vave = 2.3 m/s / (b)
vave = Δx / Δt
vave = (57.5-9.2) / (5 – 2)
vave = 16.1 m/s / (c)
vave = Δx / Δt
vave = (57.5-0) / (5 – 0)
vave = 11.5 m/s
Ch 2.2Instantaneous 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 position-time 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)
vave = Δx / Δt
vave = (6–3.5)/(0–3.6)
vave = - 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 2.3Acceleration
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/s2.)
vave = x / t½(30+0) = 2 / t
t = 2/15 seconds / a = v / t
a = 30 / (2/15)
a = 225 cm/s2
Ch 2.3Acceleration
A particle moves along the x-axis according to the equation x = 2 + 3t – t2, 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 – t2x = 2 + 3(3) – 32
x = 2 m / b)v = dx / dt
v = 3 – 2t
v = -3 m/s / c)a = d2x / dt2
a = dv/dt
a = -2 m/s2
Ch 2.4Motion 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.
/ orCh 2.51D Motion with Constant Acceleration
A particle moves along the x-axis. Its position is given by the equation x = t2 - 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 / dtv = 2t - 3
From calculus class; maximums occur when the derivative equals zero / 0 = 2t - 3
t = 1.5 seconds
x(t) = t2 - 3t + 3
x = 0.75 m / (b)
x(0) = t2 - 3t + 3
x(0) = 3 meters
3 = t2 - 3t + 3 / t2 = 3t
t = 3 sec
v(3) = 2t - 3
v(3) = 3 m/s
Ch 2.6Freely 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? (vsound = 340 m/s)
dr = ½atr2 + votr + dodr = ½ 10tr2
dr = 5 tr2 / ds = ½ats2 + vots + do
ds = vots / 5 tr2 = 340(2 - tr)
5 tr2 = 680 - 340tr
tr2 + 68tr = 136
(tr + 34)2 = 136 + (34)2
(tr + 34)2 = 1292
tr = -34 ± 35.9444
tr = 1.9444
dr = 5 tr2
dr = 18.9 meters
We know that the distance sound that travels is the same as the distance the rock travels. / Given:
vo of sound = 340 m/s
tr + ts = 2 sec
ts = 2 - tr
The rest is algebra
dr = ds
5 tr2= vots
5 tr2= 340ts
Ch 2.7Kinematics 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 ax(t), velocity vx(t), and position x(t), given that its initial acceleration, velocity, and position are axi, vxi, xi, respectively. /
Ch 2BONUS
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, vA, find the speed of B, vB, when = 80. /
Given:
A is at position +x, and get smaller as B, at +y, gets larger, thus:
-va = dx / dt
+vb = dy / dt
need to take derivative; x & y change as time progresses / x2 + y2 = L2
2x dx/dt + 2y dy/dt = 0
2x (-va) + 2y (+vb) = 0
vb = va * x/y cot = x/y
vb = va cot
vb = va cot 80
vb = 0.176 va
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 2BONUS
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?
dG = do+ vot + ½ at2: Gazelle140 = vG (6–1 s) + ½ (v / t) t2
140 = vG 5 sec + ½ v 1 sec
140 = 5.5vG See note at bottom
vG = 25.5 m/s
aG = v / t
aG = 25.5 / 1
aG = 25.5 m/s2 / dC = do+ vot + ½ at2: Cheetah
140 = vC (6 – 3 sec) + ½ (v / t) t2
140 = vC 3 sec + ½ v 3 sec
140 = 4.5vC
vC = 31.1 m/s
aC = v / t
aC = 31.1 / 3
aC = 10.4 m/s2
c) The gazelle is ahead by 11.4 m at 4 seconds
dG = ½ aG*12+ vG (4-1 sec)
dG = 89.3 m / dC = ½ aC 32 + vC (4 – 3 sec)
dC = 77.9 m
Note: At the end of the acceleration period is the maximum velocity, vG = 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 x-axis according to the equation x(t) = 3t2 – 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) = 3t2 – 2t + 3x(2) = 11 m; x(3) = 24 m
vave = Δx / Δt
vave = (24-11)/(3-2)
vave = 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 = (16-10)/(3-2)
a = 6 m/s / (d)
a = dv / dt
a = 6
a(2) = 6 m/s2
a(3) = 6 m/s2
Ch 2.5#25
A particle moves along the x-axis. Its position is given by the equation x = -4t2 + 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 / dtv = -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) = -4t2 + 3t + 2
x = 2.56 seconds / (b) x(0) = -4t2 + 3t + 2
x(0) = 2 meters
2 = -4t2 + 3t + 2
4t2 = 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 1-D such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, velocity, and position are axi, vxi, xi, respectively. (b) Show that ax2 = axi2 + 2J(vx – vxi)
J = Δa / Δtt = (af – ai) / J
Since a is not constant
aave= Δv / Δt
(af + ai)/2 = (vf – vi) / t
(af + ai) t = 2(vf – vi)
substitute in “t” from above
(af + ai) (af – ai) / J = 2(vf – vi)
(af + ai) (af – ai)= 2J(vf – vi)
af2 – ai2= 2J(vf – vi)
af2 = ai2 + 2J(vf – vi) /
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/s2 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/s2. / 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 = do+ vot + ½ at2
1000 = 0 + 80t + ½(4) t2
2t2+ 80t – 1000 = 0
t2+ 40t – 500 = 0
(t - 10 ) (t + 50 ) = 0
t = 10 seconds
(can’t equal -50 seconds)
ay= vy / t
4 = (vf – 80) / 10
vf= 120 m/s / Bvi of part B is vf of part A
in part A, a = 4 m/s2,
in part B its in free fall, a = -10 m/s2
ay= vy / t
-10 = (0 – 120) / t
t = 12 s
d = do + vot + ½ at2
d = 1000 + 120 (12) + ½(-10) 122
d = 1720 meters
CNow we have a free fall problem starting from 1720 meters
d = do + vot + ½ at2
0 = 1720 + 0 + ½ (-10) t2
t = 18.55 s
a = (vf – vi) / t
-10 = (vf – 0) / 18.55 sec
vf= 185.5 m/s
Ch 2#65
In a 100-m 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?
dM = do + vot + ½ at2: Maggie100 = vM (10.2 – 2 sec) + ½ (v / t) t2
100 = vM 8.2 sec + ½ v t
100 = vM 8.2 sec + ½ v 2 / dJ = do + vot + ½ at2: Judy
100 = vJ (10.2 – 3 sec) + ½ (v / t) t2
100 = vJ 7.2 sec + ½ v t
100 = vJ 7.2 sec + ½ v 3
b)Thus vM = v
100 = vM 8.2 sec + ½ v 2
100 = vM 8.2 sec + ½ vM 2
100 = vM 8.2 sec + vM
100 = vM 9.2 sec
vM = 10.9 m/s / a)
aM = v / t
aM = 10.9 / 2
aM = 5.43 m/s2 / b)Thus vM = v
100 = vJ 7.2 sec + ½ v 3
100 = vJ 7.2 sec + ½ vJ 3
100 = vJ 7.2 sec + 3/2 vJ
100 = vJ 8.7 sec
vJ = 11.5 m/s / a)
aJ = v / t
aJ = 11.5 / 3
aJ = 3.83 m/s2
c) Maggie is ahead by 2.75 m at 6 seconds
dM = ½ aM*22+ vM (6-2 sec)
dM = 54.5 m / dJ = ½ aJ 32 + vJ (6 – 3 sec)
dJ = 51.75 m
Ch 2.1#3
The position vs time for a certain particle moving along the x-axis 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 = (10-0) /(2-0) = 5 m/s
(b) v = Δx / Δt = (5-0) / (4-0) = 1.25 m/s
(c) v = Δx / Δt = (5-10) /(4-2) = -2.5 m/s
(d) v = Δx / Δt = (-5-5) /(7-4) = -3 1/3 m/s /
(e) v = Δx / Δt = (0-0) / (8-0) = 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 sb) t = 3 sc) t = 4.5 sd) 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 sv = x / t = (6 – 4)m / 0.4s = 5 m/s / b)t = 4.5 s
v = x / t = (5 –5)m / 0.1s = 0 m/s
c)t = 3 s
v = x / t = (7 – 8)m / 0.4s = -2.5 m/s / d)t = 7.5 s
v = x / t = (0 – -4)m / 0.8s = 5 m/s
Ch 2.51D Motion with Constant Acceleration
A particle moves along the x-axis. Its position is given by the equation x = t2 - 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 / dtv = 2t - 3
From calculus class; maximums occur when the derivative equals zero / 0 = 2t - 3
t = 1.5 seconds
x(t) = t2 - 3t + 3
x = 0.75 m / (b)
x(0) = t2 - 3t + 3
x(0) = 3 meters
3 = t2 - 3t + 3 / t2 = 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/s2 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?
aave = v / taave = (vf – vi) / 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?
vave = (vf + vi) / 2 / vave = (xf – xi) / t; where the distance, d = xf – xi(vf + vi) / 2 = (xf – xi) / 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