Question

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

(a) Find the maximum speed.

.... m/s

(b) Find the maximum acceleration of the particle.

.... m/s^{2}

(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**

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