Particle motion with a(t)=4cos(2t), v(0)=1, x(0)=0

[Blade]

A particle moves along the x-axis in such a way that its acceleration at time t for t>= 0 is given by a(t)=4cos(2t). At time t=0, the velocity of the particle is v(0)=1 and its position is x(0)=0.
a. Write an equation for the velocity of v(t) of the particle.
b. Write an equation for the position x(t) of the particle.
c. For what values of t, 0<=t<=pi, is the particle at rest

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a = dv/dt
4*cos(2t) = dv/dt
v = Integral[4*cos(2t) dt] + C
v = 2 * sin(2t) + C
t = 0 -> v = 1.
1 = 2 * sin(2*0) + C
C = 1
v(t) = 2 * sin(2t) + 1

How would I go about x(t)?

And for c. 2sin(2t)+1=0 sin2t=-1/2
arcsin2t=(1/2)
t=-.261799?

 

[himanshu121]

$$\frac{d^2x}{dt^2}=4\cos(2t)$$
$$\frac{dx}{dt}=\int 4\cos(2t)dt+c = 2sin2t+c$$
$$x=\int {2sin2t+c}dt+k$$

 

[M98Ranger]

b. To find the position x(t) of the particle, we need to integrate the velocity function with respect to time. So, we have:
x(t) = Integral[2*sin(2t) + 1 dt] + C
x(t) = -cos(2t) + t + C
t = 0 -> x(0) = 0
0 = -cos(2*0) + 0 + C
C = 1
x(t) = -cos(2t) + t + 1

c. To find when the particle is at rest, we need to find the values of t for which the velocity is equal to 0. So, we have:
v(t) = 2*sin(2t) + 1 = 0
sin(2t) = -1/2
2t = arcsin(-1/2)
2t = -pi/6 or 5pi/6
t = -pi/12 or 5pi/12

Therefore, the particle is at rest at t = -pi/12 or 5pi/12, which are both within the given range of 0<=t<=pi.