FIND WHEN PARTICLE CHANGES ITS DIRECTION

When the particle changes direction, v(t) then changes its sign.

Find distance covered in between two time :

If t_c is the time point between the time points t₁ and t₂ (t₁ < t_c 2) where the particle changes direction then the total distance travelled from time t₁ to time t₂ is calculated as

A particle moves along a line according to the law

s(t) = 2t 3 -9t 2 +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.

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

s(t) = 2t 3 -9t 2 +12t-4

v(t) = s'(t) = 6t 2 -18t+12

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

So, the particle its direction between 1 to 2.

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

s(t) = 2t 3 -9t 2 +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|

(iii) v(t) = s'(t) = 6t 2 -18t+12

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

A(1) = 12(1) - 18 ==> -6 m/sec 2

A(2) = 12(2) - 18 ==> 6 m/sec 2

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