FIND WHEN PARTICLE CHANGES ITS DIRECTION

Example 1 :

A particle moves along a line according to the law

s(t)  =  2t³-9t²+12t-4 , where t ≥ 0 .

(i) At what times the particle changes direction?

(ii) Find the total distance travelled by the particle in the first 4 seconds.

(iii) Find the particle’s acceleration each time the velocity is zero.

Solution :

Let us find when the particle is at rest. So, v(t)  =  0

s(t)  =  2t³-9t²+12t-4

v(t)  =  s'(t)  =  6t²-18t+12

6t²-18t+12  =  0

t²-3t+2  =  0

(t-1)(t-2)  =  0

t  = 1 and t  =  2

So, when t = 1 and when t = 2 the particle is at rest.

Say t < 1, t  = 0.5

v(t)  =  (t-1)(t-2)

v(0.5)  =   (0.5-1)(0.5-2)

v(0.5) > 0

Say 1 < t < 2, at t  =  1.5

v(t)  =  (t-1)(t-2)

v(1.5)  =   (1.5-1)(1.5-2)

v(1.5) < 0

So, the particle its direction between 1 to 2.

(ii)  Distance travelled by the particle in 4 seconds :

D  =  |s(0)-s(1)| + |s(1)-s(2)| + |s(2)-s(3)| + |s(3)-s(4)|

s(t)  =  2t³-9t²+12t-4

s(0)  =  -4, s(1)  =  1, s(2)  =  0, s(3)  =  5, s(4)  =  28

Applying in D, we get

D  =  |-4-1| + |1-0| + |0-5| + |5-28|

D  =  5 + 1 + 5 + 23

D  =  34 m

(iii)  v(t)  =  s'(t)  =  6t²-18t+12

A(t)  =  v'(t)  =  12t-18

The velocity becomes zero at t  =  1 and t  =  2.

A(1)  =  12(1) - 18  ==>  -6 m/sec²

A(2)  =  12(2) - 18  ==>  6 m/sec²

Example 2 :

A particle moves along a horizontal line such that its position at any time t ≥ 0 is given by

s(t)  =  t³ - 6t² + 9t + 1

where s is measured in meters and t in seconds?

(i) At what time the particle is at rest?

(ii) At what time the particle changes its direction?

(iii) Find the total distance travelled by the particle in the first 2 seconds.

Solution :

Given that s(t)  =  t³ - 6t² + 9t + 1

On differentiating s(t), we get

s'(t) = 3t² - 6(2t) + 9(1) + 0

= 3t² - 12t + 9

v(t) = 3t² - 12t + 9

On differentiating v(t), we get

v'(t) = 3(2t) - 12(1) + 0

= 6t - 12

a(t) = 6t - 12 

Let us find when the particle is at rest. So, v(t)  =  0

3t² - 12t + 9 = 0

Dividing by 3, we get

t² - 4t + 3 = 0

(t - 1)(t - 3) = 0

t = 1 and t = 3

So, when t = 1 and when t = 3 the particle is at rest.

(ii) The particle changes its direction when v (t) changes its sign. Now.

• if 0 ≤ t < 1 then both (t - 1) < 0 and (t - 3) < 0, then v(t) > 0
• if 1 < t < 3 then both (t - 1) > 0 and (t - 3) < 0, then v(t) < 0
• if t > 3 then both (t - 1) > 0 and (t - 3) > 0, then v(t) > 0

Therefore the particle changes its direction when t = 1 and t = 3.

iii)  The total distance travelled by the particle from time t = 0 to t = 2 is given by 

= |s(0) - s(1)| + |s(1) - s(2)|

= |1 - 5| + |5 - 3|

= |-4| + |2|

= 4 + 2

= 6 meters

Example 3 :

The particle moves along a straight line is such a way that after t seconds its distance from the origin is 

s(t) = 2t² + 3t meters

a) Find the average velocity between t = 3 and t = 6 seconds

b)  Find the instantaneous velocities at t = 3 and t = 6 seconds.

Solution :

To find the velocity, we find the derivative of s(t).

s(t) = 2t² + 3t

Differentiating s(t), 

s'(t) = 2(2t) + 3(1)

= 4t + 3

v(t) = 4t + 3

Average velocity = [s(b) - s(a)] / (b - a)

= (90 - 27) / (6 - 3)

= 63/3

= 21 m/sec

b)  Finding instantaneous velocity :

t = 3 and t = 6 seconds

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