1.)The position versus time for a certain particle moving along the x axis is sh
ID: 1654814 • Letter: 1
Question
1.)The position versus time for a certain particle moving along the x axis is shown in the figure below. Find the average velocity in the following time intervals.
(b) 0 to 3 s =
8.)A particle moves along the x axis according to the equation
x = 1.99 + 2.93t 1.00t2,
where x is in meters and t is in seconds.
(a) Find the position of the particle at t = 2.60 s.
m=
(b) Find its velocity at t = 2.60 s.
m/s=
(c) Find its acceleration at t = 2.60 s.
m/s2=
12.)Automotive engineers refer to the time rate of change of acceleration as the "jerk." Assume an object moves in one dimension 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 ai, vi, and xi, respectively. (Use any variable or symbol stated above as necessary.)
(b) Show that ax2 = ai2 + 2J(vx vi). (Do this on paper. Your instructor may ask you to turn in this work.)
13.)The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."
(a) Find the upward velocity with which the insect takes off.
m/s=
(b) In what time interval does it reach this velocity?
ms=
(c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.
m=
14.)A man drops a rock into a well.
(a) The man hears the sound of the splash 2.58 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water?
m=
(b) If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?
%=
Explanation / Answer
1)
at t = 0 , Xi = 0 m
at t = 3 , Xf = 7 m
D = displacement = Xf - Xi = 7 - 0 = 7 m
t = time = 3 - 0 = 3 sec
average velocity is given as
Vavg = D/t = 7/3 = 2.33 m/s
8)
x = 1.99 + 2.93 t 1.00 t2
a)
at t = 2.6
x = 1.99 + 2.93 (2.6) 1.00 (2.6)2
x = 2.85 m
b)
velocity at any time is given as
v = dx/dt = 2.93 - 2t
at t = 2.6
v = 2.93 - 2(2.6) = - 2.27 m
c)
acceleration is given as
a = dv/dt = - 2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.