Who can help me in physics (kinematic problem)?

Question

A particle displacement, S meters, from an origin O is defined by the equation S(t) = sin*4t - cos*4t at time t seconds

a. find its velocity and acceleration functions in t

b. what is the particle's max speed?

i really really really need your help!!!

Answer 1

aha! intetesting...

a. veloc = d/dt ((S)t)) = 4.cos.4t - 4.sin.4t

(is that right for d/dt of cos and sin?? school was a long time ago.

accel = d/dt veloc = 16.sin.4t-16.cos.4t

hope that's right too!

b. at max speed accel = 0 - d/dt of veloc will be + or - either side!!

so 16.sin.4t - 16.cos.4t = 0 !!

therefore sin.4t = cos.4t, but '4t' must be expressed in radians for this,

let's say sin.4t is A,

sin.4t= sqrt(1-sin.4t.sin.4t)

so A=sqrt (1-A.A)

A.A = 1-A.A ,

2.A.A = 1,

A.A = 0.5 ,

A= 0.707  = sin.4t, 4.t = 0.785 (radians)

t= 0.196 sec when max speed reached,

veloc = 4.cos.4t - 4.sin.4t = 4.0.707 - 4.0.706 = 4.0.1 = 0.4 m/sec

Answer 2

The velocity is just the derivative of position:

v(t)=d/dt(Sin4t-Cos4t)=4Cos4t+4Sin4t

Then acceleration is the derivative of velocity:  Go on, try it!

Max speed should occur when acceleration passes through a zero value (goes from negative to positive or vice versa).  So once you find the expression for a(t), set it equal to zero and solve.

Answer 3

For the velocity just take the derivative:

v = dS/dt = 4*cos(4t) + 4*sin(4t) = 4*[cos(4t) + sin(4t)]

For the acceleration, take the derivative again:

a = d^2/dt^2 = -16*sin(4t) + 16*cos(4t) = 16*[-sin(4t) + cos(4t)]

the max (or min) speed is when dv/dt = 0 (which is just a):

-16*sin(4t) + 16*cos(4t) = 0

tan(4t) = 1 which gives: 4t = pi/4 or t = pi/16

Now put this into v = cos(4t) + sin(4t) and see what you get:

v = 4*[cos(pi/4) + sin(pi/4)] = 4*[SQRT(2)/2 + SQRT(2)/2]

v = 4*SQRT(2)

v = 5.657 m/s

So the particles has a max speed of 5.657 m/s