a particle moves along a line so that at any time t its position is given by x(t)=2(pi)t + cos2(pi)t
find velocity
find acceleration
what are the values of t, 0<t<3, for which the particle is at rest?
what is the maximum velocity?
To find the velocity and acceleration of the particle, we need to differentiate the position function with respect to time.
1. Velocity (v): The velocity is the derivative of the position function x(t) with respect to time (t). So, we can find the velocity by taking the derivative of x(t).
v(t) = d/dt [x(t)]
Let's differentiate x(t) term by term:
v(t) = d/dt [2(pi)t + cos(2(pi)t)]
= 2(pi) - 2(pi)sin(2(pi)t)
2. Acceleration (a): The acceleration is the derivative of the velocity function with respect to time. Therefore, we need to differentiate the velocity function v(t) with respect to time (t).
a(t) = d/dt [v(t)]
Differentiating v(t) term by term:
a(t) = d/dt [2(pi) - 2(pi)sin(2(pi)t)]
= -4(pi)^2cos(2(pi)t)
Now, let's move on to finding the values of t (0 < t < 3) for which the particle is at rest.
3. Particle at rest: When the particle is at rest, its velocity is equal to zero. So, we need to solve the equation v(t) = 0 to find the values of t.
2(pi) - 2(pi)sin(2(pi)t) = 0
Simplifying the equation:
sin(2(pi)t) = 1
2(pi)t = pi/2
t = 1/4
So, the particle is at rest when t = 1/4.
Lastly, let's determine the maximum velocity.
4. Maximum velocity: To find the maximum velocity, we can look for the maximum value of the absolute value of the velocity function within the given interval (0 < t < 3).
To find the maximum value of a function, we need to check the critical points (where the derivative is zero) and endpoints of the interval.
The critical points of v(t) occur when sin(2(pi)t) = 1. We already found t = 1/4 as one of the critical points.
Let's check the endpoints:
When t = 0, v(t) = 2(pi).
When t = 3, v(t) = -2(pi).
Comparing the values, we can see that the maximum velocity occurs at t = 0, where v(t) = 2(pi).
Therefore, the maximum velocity is 2(pi).