Civil Engineering - Applied Mechanics - Discussion

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

Answer: Option D

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.

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