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
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.
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.
dx/dt = 3t^2 + 2t.
dx^2/dt^2 = 6t + 2.
(1)
Himanshu said:
1 decade ago
How this is possible ? the double differentiation of equation is 6t+2.
Zuber said:
1 decade ago
x = t2(t+1).
= t3+t.
dx/dt = 3t2+1.
d2x/dt2 = 6t+0.
d2x/dt2 = Acceleration.
So acceleration = 6t.
= t3+t.
dx/dt = 3t2+1.
d2x/dt2 = 6t+0.
d2x/dt2 = Acceleration.
So acceleration = 6t.
Hitesh 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 of at the any instant of time t.
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:
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.
V=dx/dt = 3t^2+2t.
a = dV/dt = 6t+2.
Answer is D.
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.
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:
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.
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:
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.
x = (t^3)+(t^2).
dx/dt = 3*t^2 + 2*t.
dx^2/dt^2 = 6t + 2.
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.
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.
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