When the particle is at rest then v(t) = 0
When the particle is moving forward then v(t) > 0
When the particle is moving backward then v(t) < 0
When the particle changes direction, v(t) then changes its sign.
Find distance covered in between two time :
If tc is the time point between the time points t1 and t2 (t1 < tc <t2) where the particle changes direction then the total distance travelled from time t1 to time t2 is calculated as
|s(t1) - s(tc)| + |s(tc) -s(t2)|
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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