# Civil Engineering - Applied Mechanics - Discussion

### Discussion :: Applied Mechanics - Section 1 (Q.No.12)

12.

A particle moves along a straight line such that distance x traversed in t seconds is given by x = t2(t + 1), the acceleration of the particle, will be

 [A]. 3t3 - 2t [B]. 3t2 + 2t [C]. 6t - 2 [D]. 6t + 2 [E]. 3t - 2

Explanation:

No answer description available for this question.

 Himanshu said: (Jun 26, 2013) How this is possible ? the double differentiation of equation is 6t+2.

 Zuber said: (Aug 27, 2013) x = t2(t+1). = t3+t. dx/dt = 3t2+1. d2x/dt2 = 6t+0. d2x/dt2 = Acceleration. So acceleration = 6t.

 Hitesh said: (Oct 4, 2013) x = t2(t+1). x = t3+t2. dx/dt = 3t2+2t. d2x/dt2 = 6t+2. This is the required answer of the acceleration of at the any instant of time t.

 Alfifi said: (Feb 13, 2014) x = t^2(t + 1)= t^3+t^2. V=dx/dt = 3t^2+2t. a = dV/dt = 6t+2. Answer is D.

 Saubhagya Nayak said: (Jun 13, 2014) If x = t^2(t+1). i.e; x = t^3+t^2. Velocity v=dx/dt. i.e; 3t^2+2t. And then acceleration a=dv/dt. i.e; 6t+2. So the ans is no doubt option D. 6t+2.

 Pinjari Shahil said: (Aug 24, 2014) x = t2(t+1). x = t3+t2. dx/dt = 3t2+2t. d2x/dt2 = 6t+2. This is the required answer of the acceleration but acceleration negative sign indicated.

 Piyush said: (Sep 12, 2014) x = t^2 (t+1). x = (t^3)+(t^2). dx/dt = 3*t^2 + 2*t. dx^2/dt^2 = 6t + 2.

 Saurabh said: (Jan 22, 2015) x = t^2(t+1). x = t^3+t^2. dx/dt = 3*t^2+2*t. d^2x/dt^2 = 6t+2. So answer is 6t+2 not 6t-2.

 Nilesh Gagare said: (Apr 27, 2017) T2(t+1) T3+t. Diff w.r.t:t 3t2+1. Again diff w.r.t:t 6t. And the answer is 6t.

 Bvc said: (Jun 16, 2017) x=t3+t2, dx/dt=3t2+2T, d2x/d2t=6T+2, Ans: D.

 Mohbat Khan Marri said: (Sep 13, 2017) t^2(t+1). =t^3+t^2, dx/dt=3t^2+2t, dx^2/dt^2=6t+2.