# 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*= 4*t*^{3}+ 3*t*^{2}- 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, isDiscussion:

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:
8 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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