Question

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

Answer #1

A particle's velocity along the x-axis is described by v(t)= At
+ Bt2, where t is in seconds, v is in m/s,
A= 0.85 m/s2, and B= -0.69 m/s3.
Acceleration= -0.53 m/s2 @ 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?

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 ?

A particle travels along a straight line with a velocity
v=(12−3t^2) m/s , where t is in seconds. When t = 1 s, the particle
is located 10 m to the left of the origin.
Determine the displacement from t = 0 to t = 7 s.
Determine the distance the particle travels during the time
period given in previous part.

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?

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 butterfly flies along with a velocity vector given by
v = (a-bt²) Î + (ct) ĵ where a=1.4 m/s, b=6.2 m/s³, and c=2.2 m/s².
When t= 0 seconds, the butterfly is located at the origin.
Calculate the butterfly’s position vector and acceleration vector
as functions of time. What is the y-coordinate as it flies over x =
0 meters after t = 0 seconds?”

The velocity of a particle moving along a line is a function of
time given by v(t)=81/(t2+9t+18). Find the
distance
that the particle has traveled after t=9 seconds if it started at
t=0 seconds.

A particle moves along a line with velocity v(t)=(3 -
t)(2+t), find the distance traveled during the time interval [0,
1].

The velocity of a particle constrained to move along the x-axis
as a function of time t is given by:
v(t)v(t)=−(−(14/t0)sin(t/t0)/t0)sin(t/t0).
Part A:
If the particle is at x=8 m when t=0, what is its position at t
= 9t0. You will not need the value of t0 to solve any part of this
problem. If it is bothering you, feel free to set
t0=1everywhere.
Part B:
Denote instantaneous acceleration of this particle by a(t).
Evaluate the expression 8 +v(0)t+a(0)t2/2+v(0)t+a(0)t2/2...

Practice Derivatives and integrals. A particle’s velocity is
described by the function v = ( t^2 – 7t + 10) m/s, where t is in
s.
a) Graph the velocity function for t in the interval 0s-6s.
b) At what times does the particle reach its turning points?
c) Find and graph the position function x (t).
d) Find and graph the acceleration function a(t).
e) What is the particle’s acceleration at each of the turning
points?

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