HOW TO KNOW WHEN A PARTICLE CHANGES DIRECTION CALCULUS

Problem 1 :

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 :

s(t) = t³ − 6t² + 9t + 1

(i) When the particle is at rest, its velocity will become 0.

s(t) = t³ − 6t² + 9t + 1

v(t) = 3t² - 12t + 9 + 0

0 = 3t² - 12t + 9

3(t² - 4t + 3) = 0

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

t = 1 and t = 3

(ii) 

If 0 ≤ t < 1

v(t) = (t - 1)(t - 3)

say t = 0.5, then v(t) = + > 0

If 1 ≤ t < 3

say t = 2, then v(t) = - < 0

If t > 3

say t = 4, then v(t) = + > 0

Then the particle changes its direction in between 1 to 3 seconds.

(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|

= 6 meters

Problem 2 :

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 :

 When the velocity will become 0, 

s(t) = 2t³ − 9t² +12t − 4

v(t) = 6t² - 18t + 12 - 0

v(t) = 0

6(t² - 3t + 2) = 0

(t - 1)(t - 2) = 0 

t = 1 and t = 2

(i) 

If 0 ≤ t < 1

v(t) = (t - 1)(t - 2)

say t = 0.5, then v(t) = + > 0

If 1 ≤ t < 2

say t = 1.5, then v(t) = - < 0

If t > 2

say t = 3, then v(t) = + > 0

Then the particle changes its direction in between 1 to 2 seconds.

(iii)  The total distance travelled by the particle from time t = 0 to t = 4 is given by,

= |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) = 2 - 9 + 12 - 4 = 1

s(2) = 16 - 36 + 24 - 4 = 0

s(3) = 54 − 81 + 36 − 4

= 90 - 85

= 5

s(4) = 128 - 144 + 48 - 4

= 176 - 148

= 28

=|-4 - 1| + |1 - 0| + |0 - 5| + |5 - 28|

= |-5| + |1| + |-5| + |-23|

= 5 + 1 + 5 + 23

= 34 meters

Total distance covered in first 4 seconds is 34 meters.

(iii) v(t) = 6t² - 18t + 12

a(t) = 12t - 18

At t = 1 and t = 2, the velocity will become 0.

a(1) = 12(1) - 18 ==> -6 m/sec²

a(2) = 12(2) - 18 ==> 6 m/sec²