Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 8)
8.
The displacement of a particle which moves along a straight line is given by S = 4t3 + 3t2 - 10 where S is in meters and t is in seconds. The time taken by the particle to acquire a velocity of 18 m/sec from rest, is
sec1 sec
1.5 sec
1.5 sec.
Discussion:
2 comments Page 1 of 1.
Masroor Ali said:
3 years ago
Simply dx/dt=12t^2 + 6t.
As v=18m/s so putting t = 1s.
18m/s=12 (1) +6 (1).
L. H. S=R. H. S.
As v=18m/s so putting t = 1s.
18m/s=12 (1) +6 (1).
L. H. S=R. H. S.
Rohit kumar said:
9 years ago
let ds/dt = 12t sqa + 6t - 18 = 0(By Differentation)
Now, 2t Squa + t - 3 = 0.
Hence, 2tsqua - 2t + 3t - 3 = 0.
OR
2t(t - 1) + 3(t - 1) =0.
So (2t + 3) (t - 1) = 0 -> 2t + 3=0 And t - 1=0.
So t = 1 (Because 2t = -3 And t = -1.5 which is not possiable).
Now, 2t Squa + t - 3 = 0.
Hence, 2tsqua - 2t + 3t - 3 = 0.
OR
2t(t - 1) + 3(t - 1) =0.
So (2t + 3) (t - 1) = 0 -> 2t + 3=0 And t - 1=0.
So t = 1 (Because 2t = -3 And t = -1.5 which is not possiable).
(1)
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