Question

A particle starts at the origin with initial velocity ⃗v(0) = ⃗i − ⃗j + ⃗k. Its acceleration is ⃗a(t) = 4t⃗i + 3t⃗j − ⃗k. Find its position at t = 3.

Answer #1

A moving particle starts at an initial position
r(0) = <1, 0, 0> with initial velocity
v(0) = i - j +
k. Its acceleration is a(t) = 4t
i + 4t j +
k.
Find its velocity, v(t), and position,
r(t), at time t.

Find the position vector of a particle that has acceleration
2i+4tj+3t^2k, initial velocity v(0)=j+k and initial position
r(0)=j+k

Given that the acceleration vector is a(t)=(-9 cos(3t))i+(-9
sin(3t))j+(-5t)k, the initial velocity is v(0)=i+k, and the initial
position vector is r(0)=i+j+k, compute:
A. The velocity vector v(t)
B. The position vector r(t)

please ASAP!!
Suppose that a particle has the following acceleration vector
and initial velocity and position vectors.
a(t) = 5 i +
9t k,
v(0) = 3 i
−
j, r(0)
= j + 6 k
(a)
Find the velocity of the particle at time t.
(b)
Find the position of the particle at time t.

Consider a particle moving through space with velocity v(t) =
cos(t)i−sin(t)j + tk.
(i) Determine its acceleration vector.
(ii) Determine the position vector, supposing the
particle starts at position (3,−2,3) at time t = 0.

A particle starts from the origin with velocity 5 ?̂m/s at t = 0
and moves in the xy plane with a varying acceleration given by ?⃗ =
(2? ?̂+ 6√? ?̂), where ?⃗ is in meters per second squared and t is
in seconds. i) Determine the VELOCITY and the POSITION of the
particle as a function of time.

A) A particle starts from the origin with velocity 5 ?̂m/s at t
= 0 and moves in the xy plane with a varying acceleration given by
?⃗ = (2? ?̂+ 6√? ?̂), where ?⃗ is in meters per second squared and
t is in seconds.
i) Determine the velocity of the particle as a function of
time.
ii) Determine the position of the particle as a function of
time.
(Explanation please )

Find the velocity and position vectors of a particle that has
the given acceleration and the given initial velocity and
position.
a(t) = 2 i +
6t j + 12t2
k, v(0) = i,
r(0) = 3 j − 6
k

A 3.00-kg particle starts from the origin at time zero. Its
velocity as a function of time is given by v = (3t^2) i+ (2t) j
where v is in meters per second and t is in seconds.
(a) Find its position at t = 1s.
(b) What is its acceleration at t = 1s ?
(c) What is the net force exerted on the particle at t = 1s
?
(d) What is the net torque about the origin...

When at the origin (0,0) the initial velocity of a particle is
?⃗0 = (2.0?̂ + 5.0?̂) ? / ? and its constant acceleration is ?⃗ =
(7.0?̂ + 5.0?̂) ? / ? two . Find the final position of the particle
when its velocity is ?⃗ = (16.0?̂ + 15.0?̂) ? / ?. (First find the
time it takes to reach your position final).

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