Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 4)
4.
The equation of motion of a particle starting from rest along a straight line is x = t3 - 3t2 + 5. The ratio of the accelerations after 5 sec and 3 sec will be
Discussion:
14 comments Page 1 of 2.
Suman ghimire said:
3 years ago
Equation of motion x = t^3-3t^2+5.
The equation to calculate the distance (in question it is given that the particle is moving along a straight line).
Differentiate with time to get velocity:
v = dx/dt = 3t^2-6t --> Velocity.
a = d^2x/dt = 6t-6 --> Acceleration.
Hence, when t = 5 sec.
a = 6*5 - 6 = 24.
When t = 3sec.
a = 6*3 - 6 = 12.
Ratio = 24/2 = 2.
The equation to calculate the distance (in question it is given that the particle is moving along a straight line).
Differentiate with time to get velocity:
v = dx/dt = 3t^2-6t --> Velocity.
a = d^2x/dt = 6t-6 --> Acceleration.
Hence, when t = 5 sec.
a = 6*5 - 6 = 24.
When t = 3sec.
a = 6*3 - 6 = 12.
Ratio = 24/2 = 2.
(14)
Engr.Mohbat Baloch said:
8 years ago
x = t3 - 3t2 + 5.
dx/dt = t2/2 - 3t.
putting t=5
= 12.5 - 15.
= -2.5.
putting t=3
4.5 - 9.
= -4.5.
4.5/2.5 = 1.8
~2.
dx/dt = t2/2 - 3t.
putting t=5
= 12.5 - 15.
= -2.5.
putting t=3
4.5 - 9.
= -4.5.
4.5/2.5 = 1.8
~2.
(2)
Jadecliff said:
1 decade ago
x = Distance.
v = Velocity = dx/dt.
a = Acceleration = dv/dt.
From x = t3 - 3t2 + 5.
v = 3t2 - 6t.
a = 6t - 6.
By substituting 5 sec and 3 sec in a.
a1 = 24.
a2 = 12.
a1 : a2 = 2 : 1.
v = Velocity = dx/dt.
a = Acceleration = dv/dt.
From x = t3 - 3t2 + 5.
v = 3t2 - 6t.
a = 6t - 6.
By substituting 5 sec and 3 sec in a.
a1 = 24.
a2 = 12.
a1 : a2 = 2 : 1.
(1)
SEKHAR ALVAKONDA said:
7 years ago
Thank you all.
(1)
Saurabh said:
1 decade ago
x = t3 - 3t2 + 5.
dx/dt = t2/2 - 3t.
= 12.5 - 15.
= -2.5.
4.5 - 9.
= -4.5.
4.5/2.5 = 1.8 = 2.
dx/dt = t2/2 - 3t.
= 12.5 - 15.
= -2.5.
4.5 - 9.
= -4.5.
4.5/2.5 = 1.8 = 2.
Dawit tamiru said:
1 decade ago
Another method d"x/dt = 6t-6.
6(5)-6 = 24 and 6(3)-6 = 12. we got the ratio 2.
6(5)-6 = 24 and 6(3)-6 = 12. we got the ratio 2.
Priyanka marmat said:
1 decade ago
d"x/dt = 6t-6.
6(5)-6 = 24.
6(4)-6 = 12.
24/12 = 2.
The ratio is 2.
6(5)-6 = 24.
6(4)-6 = 12.
24/12 = 2.
The ratio is 2.
Abhay said:
10 years ago
Equation of motion x = t^3-3t^2+5.
Equation to calculate distance (in question it is given that the particle is moving along a straight line).
Differentiate with time to get velocity:
v = dx/dt = 3t^2-6t.....Velocity.
a = d^2x/dt = 6t-6.....Acceleration.
Hence, when t = 5 sec.
a = 6*5 - 6 = 24.
When t = 3sec.
a = 6*3 - 6 = 12.
Ratio = 24/2 = 2.
Equation to calculate distance (in question it is given that the particle is moving along a straight line).
Differentiate with time to get velocity:
v = dx/dt = 3t^2-6t.....Velocity.
a = d^2x/dt = 6t-6.....Acceleration.
Hence, when t = 5 sec.
a = 6*5 - 6 = 24.
When t = 3sec.
a = 6*3 - 6 = 12.
Ratio = 24/2 = 2.
Tariku Teshome said:
9 years ago
Answer for;
d = x = distance.
v = velocity.
a = acceleration.
11, 5 & 2 respectively.
d = x = distance.
v = velocity.
a = acceleration.
11, 5 & 2 respectively.
Laxmipriyadhal12345@gmail said:
9 years ago
Thank you @Abhay.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers