A particle moves according to a law of motion below 0 <= t < 12, where t is measured in seconds and s in feet

Question

A particle moves according to a law of motion below 0 <= t < 12, where t is measured in seconds and s in feet.

 

s= f(t) = cos(tpi/6)

 

 

(a) Find the total distance tranvelled during the first 12 seconds.

(b) When is the particle speeding up?

(c) When is it slowing down?

Answer 1

Between (t*pi/6)=0 and (t*pi/6) = pi, the cosine function goes from positive 1 to -1.  So the distance covered is 2.  Between (t*pi/6) = pi and (t*pi/6)=2*pi, the cosine function goes from -1 to positive 1.  That distance is also two.  But what is t when the cosine function moves a total of 4 units?

 

t*pi/6 = 0 -->  t = 0

t*pi/6 = pi --> t = 6

t*pi/6 = 2*pi --> t = 12

 

So, between t = 0 and t = 12, the particle has moved 4 units.  That's part (a).

 

For parts (b) and (c), we need to know the following:  The particle is speeding up when velocity and acceleration have the same signs.  It is slowing down when they have opposite signs.

 

When 0<t<6, the velocity is negative (1st derivative).  When  6<t<12, the velocity is positive.

To find the second derivative, just look at concavity of the cosine function.

 

When 0<t<3, the function is concave down (CD).  When 3<t<9, the graph is concave up (CU).  For 9<t<12, the function is CD again.  This corresponds to the signs of the 2nd derivative (- for CD, + for CU).

 

Question (b) asks of us to find the intervals on which the first and second derivatives have the same sign.

On (0,3) they are both negative.  On (6,9) they are both positive.  The particle is speeding up on these intervals.

 

Whatever is left over in the interval [0, 12] (minus endpoints) must be where velocity and acceleration have opposite signs.  So the particle is slowing down on (3,6) and (9,12).