Question

A particle's velocity along the x-axis is described by v(t)= At
+ Bt^{2}, where t is in seconds, v is in m/s,

A= 0.85 m/s^{2}, and B= -0.69 m/s^{3}.
Acceleration= -0.53 m/s^{2} @ t=0 and the Displacement=
-2.58 m b/w t=1s to t=3s.

What is the distance traveled in meters, by the particle b/w times t=1s and t=3s?

Answer #1

A particles velocity along the x-axis is described by v(t) = At
+ Bt^2, where t is in seconds, velocity is in m/s^2, A = 1.18m/s^2
and B = -0.61m/s^3. What is the distance traveled, in m, by the
particle between times t0=1.0 and t1=3.0? please show steps and
calculations

The equation x(t) = −bt2 +
ct3 gives the position of a particle traveling
along the x axis at any time. In this expression,
b = 4.00 m/s2, c = 4.80
m/s3, and x is in meters when t is
entered in seconds. For this particle, determine the following.
(Indicate the direction with the sign of your answer as
applicable.)
(a) displacement and distance traveled during the time interval
t = 0 to t = 3 s
displacement
distance
(b)...

The velocity of a particle moving along the x-axis
varies with time according to
v(t) = A +
Bt−1,
where
A = 7 m/s,
B = 0.33 m,
and
1.0 s ≤ t ≤ 8.0 s.
Determine the acceleration (in m/s2) and position (in
m) of the particle at
t = 2.6 s
and
t = 5.6 s.
Assume that
x(t = 1 s) = 0.
t = 2.6 s
acceleration m/s2 position m
?
t = 5.6 s
acceleration m/s2
position m ?

An object moves along the x axis according to the
equation
x = 4.00t2 −
2.00t + 3.00,
where x is in meters and t is in seconds.
(a) Determine the average speed between t = 1.60 s and
t = 3.20 s.
The average speed is the distance traveled divided by the time. Is
the distance traveled equal to the displacement in this case?
m/s
(b) Determine the instantaneous speed at t = 1.60 s.
m/s
Determine the instantaneous...

A particle is launched from the origin along the
x-axis at an initial velocity of 2 m/s. If the particle is
accelerated according to the formula a(t) = -sin(t) where t is in
seconds, what is the particle's position at time t = pi
seconds?

) A particle is moving according to the velocity equation v(t) =
9t^2-8t-2 . The equation uses units of meters and seconds
appropriately. At t = 1 s the particle is located at x = 2 m. (a)
What is the particle's position at t = 2 s? (b) What is the
particle's acceleration at t = 1 s? (c) What is the particle's
average velocity from t = 2 s to t = 3 s?

The velocity-time graph of a particle moving along the x-axis is
shown. The particle has zero velocity at t = 0.00 s and reaches a
maximum velocity, vmax, after a total
elapsed time, ttotal. If the initial
position of the particle is x0 = 7.29 m, the
maximum velocity of the particle is
vmax = 11.3 m/s, and the total elapsed
time is ttotal = 25.0 s, what is the
particle's position at t = 16.7 s?
b. At t...

physics
please explain as much as possible!
Object moves following the path x=At^3-Bt^2x=At3−Bt2 , where A=1
m/s3 and B=1 m/s2. In the next questions
discuss the character of this motion in the time interval between
t=0 and t=1s. find the velocity and acceleration, and
determine if the object is at the origin. for the
following:
(a) at time t=0
(b) at time t =1/3s
(c) at time 1s

A particle travels along the path defined by the parabola
y=0.2x^2. If the component of velocity along t he x axis is
Vx=(2.9t)ft/s, where t is in seconds. determine the magnitude of
the particle's acceleration when t = 1s. when t = 0 , x =0 and y =
0.

Two particles move along an x axis. The position of
particle 1 is given by x = 10.0t2 +
6.00t + 4.00 (in meters and seconds); the acceleration of
particle 2 is given by a = -9.00t (in meters per seconds
squared and seconds) and, at t = 0, its velocity is 24.0
m/s. When the velocities of the particles match, what is their
velocity?

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