1. A jetliner, traveling northward, is landing with a speed of 75.8 m/s. Once the jet touches down, it
has 693 m of runway in which to reduce its speed to 12.6 m/s.
a. Compute the magnitude of the average acceleration of the jetliner. (5 points)
Formula: vf2 = vi2 + 2*a*d
12.62 = 75.82 + 2*a*d
158.76 - 5745.64 = 1386*a
-5586.88 = 1386*a
-5586.88/1386 = a
a = -4.031 m/s^2
b. What is the direction of the acceleration of the jet during the landing in part a above? (2 points)
the acceleration is in the negative direction. this is because the velocity is decreasing
c. If the jet continues with the acceleration determined in part a above, how much time does it take
to stop from when it touches down? (5 points)
1. Vf = vi + at
2. 0 = 75.8+(-4.031)t
3. t = 18.80 s
Answer: a. ___-4.031m/s____ b. ___negative____ c. ____18.80s____
1
, 2. A student releases a block of mass m from rest at the top of the slide h1 shown in the drawing
below. The block moves down the slide to point P and then off the end of a table of height h2,
landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance
are negligible.
a. Derive an equation for d in terms of h1, h2, m and physical constants, as appropriate. (10 points)
the starting point: Em = U = m * g * h1
the ending point: Emf = K + U = ½ * m*v2+ mgh2
there is no friction: mgh1 = ½ m*v2*mgh2
v2 = 2*g(h1-h2) ! speed
distance traveled:
x = v0x t
y = t -1/2 g*t2
height of the table: y=h2
vertical speed: t = sqrt((2h2)/g)
substitution:
d = sqrt((2h2)/g)
d = sqrt((4h2)*( h1-h2))
d = sqrt((4h2)*( H-2*h2))
b. If the experimental setup above were transported to the Moon and the experiment were
conducted on the Moon(where the acceleration of gravity is less than that on Earth) will the new
landing distance d be greater than, less than or the same as that on Earth?
___________ Greater than ___________ Less than ____ _______ The same
as (1 point)
Briefly explain your answer.
(4 points)
Gravity depends on mass, in both cases the mass is the same
2
has 693 m of runway in which to reduce its speed to 12.6 m/s.
a. Compute the magnitude of the average acceleration of the jetliner. (5 points)
Formula: vf2 = vi2 + 2*a*d
12.62 = 75.82 + 2*a*d
158.76 - 5745.64 = 1386*a
-5586.88 = 1386*a
-5586.88/1386 = a
a = -4.031 m/s^2
b. What is the direction of the acceleration of the jet during the landing in part a above? (2 points)
the acceleration is in the negative direction. this is because the velocity is decreasing
c. If the jet continues with the acceleration determined in part a above, how much time does it take
to stop from when it touches down? (5 points)
1. Vf = vi + at
2. 0 = 75.8+(-4.031)t
3. t = 18.80 s
Answer: a. ___-4.031m/s____ b. ___negative____ c. ____18.80s____
1
, 2. A student releases a block of mass m from rest at the top of the slide h1 shown in the drawing
below. The block moves down the slide to point P and then off the end of a table of height h2,
landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance
are negligible.
a. Derive an equation for d in terms of h1, h2, m and physical constants, as appropriate. (10 points)
the starting point: Em = U = m * g * h1
the ending point: Emf = K + U = ½ * m*v2+ mgh2
there is no friction: mgh1 = ½ m*v2*mgh2
v2 = 2*g(h1-h2) ! speed
distance traveled:
x = v0x t
y = t -1/2 g*t2
height of the table: y=h2
vertical speed: t = sqrt((2h2)/g)
substitution:
d = sqrt((2h2)/g)
d = sqrt((4h2)*( h1-h2))
d = sqrt((4h2)*( H-2*h2))
b. If the experimental setup above were transported to the Moon and the experiment were
conducted on the Moon(where the acceleration of gravity is less than that on Earth) will the new
landing distance d be greater than, less than or the same as that on Earth?
___________ Greater than ___________ Less than ____ _______ The same
as (1 point)
Briefly explain your answer.
(4 points)
Gravity depends on mass, in both cases the mass is the same
2
