Question

The position of a particle is given in cm by x = (9) cos 5πt, where t is in seconds.

(a) Find the maximum speed.
.... m/s

(b) Find the maximum acceleration of the particle.
.... m/s²

(c) What is the first time that the particle is at x = 0 and moving in the +x direction?
..... s

Answer 1

x = 9(cos 5πt)

a)
We know, Velocity = dx/dt.
Differentiating the given function,

velocity v = dx/dt
velocity v = d/dt (9(cos 5πt))
velocity v = - 45π (sin 5πt)

Speed will be max when value of (sin 5πt = - 1)
5πt = arc sin(-1) = 3π/2
Solving for "t",
t = 3/10s

Substituting Value -
velocity v = - 45π (sin 5π * 3/10)
Maximum Velocity v = 45 * 3.14 m/s
Maximum Velocity v = 141.37 cm/s
Maximum Velocity v = 1.41 m/s

b)
We know, acceleration = dv/dt.
Differentiating velocity function,
acceleration a = dv/dt = d/dt (- 45π (sin 5πt))
a = -225π^2 (cos 5πt)
and the maximum acceleration is when cos 5πt = -1.
Therefore,

Max Acceleration = 225 * 3.14^2 cm/s^2
Max Acceleration a_max = 2218.4 cm/s^2
Max Acceleration a_max = 22.18 m/s^2

c)
Time taken to come at this position = 3T/4
T = 2π/ w
Where w = 5π
T = 2/5
Value of 3T/4 = 3*2/4*5

t = 0.3 s
First time that the particle is at x = 0 and moving in the +x direction t = 0.3 s