Civil Engineering - Applied Mechanics - Discussion

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

[A]. 2
[B]. 3
[C]. 4
[D]. 5

Answer: Option A

Explanation:

No answer description available for this question.

Saurabh said: (Jul 22, 2013)  
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.

Jadecliff said: (Jul 23, 2013)  
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.

Dawit Tamiru said: (Nov 20, 2014)  
Another method d"x/dt = 6t-6.

6(5)-6 = 24 and 6(3)-6 = 12. we got the ratio 2.

Priyanka Marmat said: (Feb 16, 2015)  
d"x/dt = 6t-6.
6(5)-6 = 24.
6(4)-6 = 12.
24/12 = 2.

The ratio is 2.

Abhay said: (Jan 25, 2016)  
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.

Tariku Teshome said: (Jul 18, 2016)  
Answer for;

d = x = distance.
v = velocity.
a = acceleration.

11, 5 & 2 respectively.

Laxmipriyadhal12345@Gmail said: (Nov 11, 2016)  
Thank you @Abhay.

Abhinandan said: (Apr 23, 2017)  
@Jadecliff and @Abhay.

Thanks for giving the step by step solution.

Engr.Mohbat Baloch said: (Sep 10, 2017)  
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.

Eric said: (Oct 16, 2018)  
Thanks @Abhay.

It is easy to understand it.

Sekhar Alvakonda said: (Dec 14, 2018)  
Thank you all.

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