A particle moves along a coordinate axis in such a way that its position at time t is given by
s(t)=(√(3t))+2cos(t) for 0≤t≤2π . At what time is the particle at rest?
To find the time at which the particle is at rest, we need to determine when the velocity of the particle is zero.
The velocity of the particle is the derivative of its position function with respect to time. Therefore, let's find the derivative of s(t):
s(t) = √(3t) + 2cos(t)
Taking the derivative with respect to t:
s'(t) = (1/2)√(3/t) - 2sin(t)
To find when the particle is at rest, we set the velocity equal to zero and solve for t:
0 = (1/2)√(3/t) - 2sin(t)
Now, let's solve this equation for t.
To find the time at which the particle is at rest, we need to find the values of t for which the velocity of the particle is zero.
The velocity of a particle is the derivative of its position with respect to time. Therefore, we need to find the derivative of the position function s(t) with respect to t:
s(t) = √(3t) + 2cos(t)
To find the velocity, differentiate both terms of s(t) with respect to t:
s'(t) = d/dt (√(3t)) + d/dt (2cos(t))
Using the Chain Rule, we can differentiate each term separately:
s'(t) = (1/2√(3t))(d/dt (3t)) + (-2sin(t))
Simplifying, we get:
s'(t) = (1/2√(3t))(3) + (-2sin(t))
s'(t) = 3/(2√(3t)) - 2sin(t)
Now, we need to find the values of t for which the velocity s'(t) is zero. In other words, when does the equation s'(t) = 0 hold?
0 = 3/(2√(3t)) - 2sin(t)
To solve this equation, we can isolate the sin(t) term:
2sin(t) = 3/(2√(3t))
Sin(t) = 3/(4√(3t))
Now, we can solve for t. Take the inverse sine (arcsin) of both sides:
t = arcsin(3/(4√(3t)))
Unfortunately, this equation is non-linear and cannot be solved algebraically. However, we can find approximate solutions using numerical methods or graphing the equation.
it is at rest when s'(t) = 0 . That is, when
3/(2√(3t)) - 2sin(t) = 0
t = 0.5957, 2.8838