Consider a particle moving along the x-axis where x(t) is the position of the particle at time t
A particle, initially at rest, moves along the x-axis such that its acceleration at time t 0 is given by a(t)=5cos(t). At t=0, its position is x=2?

Question

(a) Find the velocity and position functions for the particle.

v(t)=
f(t)=

(b) Find the values of t for which the particle is at rest. (Use k as an arbitrary non-negative integer.)

t=

Cameron Alexander · Accepted Answer

# v=5sint #
# x=-5cost+7#

At rest when # t = npi# where #n in NN# #a=(dv)/dt and v=dx/dt # so with #a(t)=5cost# we have: #\ \ \ \ \ (dv)/dt = 5cost# #:. v=5sint+A# #v=0# when #t=0# (initially at rest); so # \ \ \ \ \ 0 = 0+A => A=0# # :. v=5sint # So then: #\ \ \ \ \ (dx)/dt = 5sint# #:. x=-5cost+B# #x=2# when #t=0#; so # \ \ \ \ \ 2 = -5+B => B=7# #:. x=-5cost+7# If the particle is at rest then #v=0# # :. 5sint = 0# # :. sint = 0# # :. t = npi# where #n in NN# Note: Some tutors and texts combine the initial conditions into a definite integral and remove the need to evaluate the constant of integration, as follows: e.g for the velocity we have: # (dv)/dt = 5cost => int (dv)/dt dt=int5costdt# If we combine the initial conditions into a definite integral then the integration variable #t# is arbitrary, then we can write # int_0^v (dv)/dt dt=int_0^t 5costdt# # :. int_0^v dv=int_0^t 5costdt# (or if you prefer #int_0^v d rho=int_0^t 5costau d tau#) # :. [v]_0^v= [5sint]_0^t# # :. v-0= 5sint - 5sin0# # :. v= 5sint #, as before Similarly for the displacement we could write # int_2^x dx= int_0^t 5sint #, as before and again we get the same solution.

Ezra Adams · Answer

undefined To find the position function $ x(t) $ of the particle, we'll integrate the given acceleration function $ a(t) = 5 \cos(t) $ twice with respect to time to obtain the position function.

Given:
Initial velocity $ v(0) = 0 $ (particle is initially at rest)
Initial position $ x(0) = 2 $

We start by integrating the acceleration function to find the velocity function:
$$ v(t) = \int a(t) \, dt = \int 5\cos(t) \, dt = 5\sin(t) + C_1 $$

Using the initial condition $ v(0) = 0 $, we find $ C_1 = 0 $, so the velocity function becomes:
$$ v(t) = 5\sin(t) $$

Next, we integrate the velocity function to find the position function:
$$ x(t) = \int v(t) \, dt = \int 5\sin(t) \, dt = -5\cos(t) + C_2 $$

Using the initial condition $ x(0) = 2 $, we find $ C_2 = 2 $, so the position function becomes:
$$ x(t) = -5\cos(t) + 2 $$

Therefore, the position function of the particle moving along the x-axis is:
$$ x(t) = -5\cos(t) + 2 $$

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t A particle, initially at rest, moves along the x-axis such that its acceleration at time t &gt; 0 is given by a(t)=5cos(t). At t=0, its position is x=2?

(a) Find the velocity and position functions for the particle. v(t)= f(t)= (b) Find the values of t for which the particle is at rest. (Use k as an arbitrary non-negative integer.) t=

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t A particle, initially at rest, moves along the x-axis such that its acceleration at time t > 0 is given by a(t)=5cos(t). At t=0, its position is x=2?

(a) Find the velocity and position functions for the particle.

v(t)=
f(t)=

(b) Find the values of t for which the particle is at rest. (Use k as an arbitrary non-negative integer.)

t=