Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : 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
3t3 - 2t
3t2 + 2t
6t - 2
6t + 2
3t - 2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Narendra Bishwokarma said:   4 years ago
Given:

x=t^2(t+1)=t^3+t^2.
Differentiating on both side
w.r.t. t => dx/dt=3t^2+2t then double differentiation: d^2x/dt^2=6t+2.
(3)

Shobhit pal said:   5 years ago
X = t^3 + t^2.
dx/dt = 3t^2 + 2t.
dx^2/dt^2 = 6t + 2.
(1)

Mohbat Khan Marri said:   8 years ago
t^2(t+1).
=t^3+t^2,
dx/dt=3t^2+2t,
dx^2/dt^2=6t+2.

BVC said:   8 years ago
x=t3+t2,
dx/dt=3t2+2T,
d2x/d2t=6T+2,
Ans: D.

Nilesh gagare said:   8 years ago
T2(t+1)
T3+t.

Diff w.r.t:t
3t2+1.
Again diff w.r.t:t
6t.

And the answer is 6t.

Saurabh said:   1 decade ago
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.

Piyush said:   1 decade ago
x = t^2 (t+1).
x = (t^3)+(t^2).
dx/dt = 3*t^2 + 2*t.
dx^2/dt^2 = 6t + 2.

Pinjari shahil said:   1 decade ago
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.

Saubhagya Nayak said:   1 decade ago
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.

ALFIFI said:   1 decade ago
x = t^2(t + 1)= t^3+t^2.
V=dx/dt = 3t^2+2t.
a = dV/dt = 6t+2.

Answer is D.


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