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

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