PHY 131
Ch 2-5 Exam
Name
A particle moves along the x-axis. Its position is given by the equation x = -t3/3 + 4t + 3 with SI units.
Determine (a) its position when it changes direction (b) its velocity after it returns to the position it had at
t = 0, and (c) its acceleration at this time.
(a)
t = ±2 seconds
(b)
x(0) = -t3/3 + 4t + 3
(c)
3
v = dx / dt
x(2) = -t /3 + 4t + 3
x(0) = 3 meters 5/5
a = d v / dt 2/2
v = -t2 + 4
x = 8.33 meters 5/5
a = d(-t2 + 4) / dt
0 = -t2 + 4
3 = -t3/3 + 4t + 3 5/5
a = -2t 3/3
3
2/2
3/3
x(-2) = -t /3 + 4t + 3
t = 0, -3.46, +3.46 3/3
a = -2(3.46)
maximums occur when
3/3
x = -2.33 meters 3/3
a = -6.93 m/s2
the derivative = 0
2
v(3.46) = -t + 4 = -8 m/s
A ball is thrown up onto a roof, landing 4.00 s later at height h = 20.0 m
above the release level. The ball's path just before landing is angled at θ =
60.0˚ with the roof. (a) Find the horizontal distance d it travels. What are
the (b) magnitude and (c) angle (relative to the horizontal) of the ball's
initial velocity?
WileyPlus hint…do the reverse problem
y = ½ a t2 + voy t + yo
0 = ½(-10)t2 + sinθv t + 20
0 = -5(4)2 + sin60 v (4) + 20
v = 17.3 m/s or vo_y = 15 m/s
11/11
(a) 4/4
vave = Δx / Δt
8.66 = d / 4
d = 34.6 meters
(b) 6/6
a = Δvy / Δt
-10 = vfy – 15 / 4
vf_y = -25 m/s
(c) 5/5
tan β = vyf / vxf
tan β = 25/8.66
β = 70.9°
v = (252 + 8.662)½
v = 26.5 m/s 3/3
solve for vx = cos60(17.3)
vx = 8.66 m/s 5/5
Two objects are connected by a light string that
passes over a frictionless pulley. If the incline is
frictionless and if m1= 5.00 kg, m2 = 7.00 kg, m3 =
8.00 kg, and  = 30.0, find the tension in the string
(a) at point A, (b) at point B, and (c) the speed of
the blocks 2.00 s after being released from rest.
m3g causes pulleys to move in CW rotation
m1g causes pulleys to move in CCW rotation
FNet
= mT
a
80N – sin30(50) = (5+7+8) a
3/3
5/5
4/4
a = 2.75 m/s2
(a)
3/3
(b)
Check
T1 +
m2(a)
38.8 + 7(2.75)
58 N
= m3(g-a)
= 58
= 58 N
3/3
2/2 2/2
FW_A = m(g sinθ + a)
FW_A = m(10 ½ + 2.75)
FW_A = 38.8 N = FT_A
2/2
2/2
2/2
FW_B = m (g - a)
FW_B = 8 (10 – 2.75)
FW_B = 58N = FT_B
(c)
a = Δvy / Δt
2.75 = (vf – 0) / 2
vf = 5.5 m/s
5/5