Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 14 (Q.No. 20)
20.
The displacement of a particle undergoing rectilinear motion along the x-axis is given by x = 2t3- 21t2 + 60t + 6 (m)
The acceleration of the particle when its velocity is zero, is
36 m/s2
± 18 m/s2
9 m/s2
- 9 m/s2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
1 comments Page 1 of 1.

Taj Mir Khan said:   1 year ago
The function's first derivative will give velocity for which we will put it = 0; solving the quadratic equation will give 2,5.

Now, take the double derivative of the function which will accelerate. Putting the values from the first derivative will give + & - 18.

i.e. dx/dt = 6t^2 - 42t +60
Now dx/dt = 0 and solve t^2 - 7t +10 will give 2,5.
Now take dx/dt^2 = 12t-42 which is "a".
Put t=2,5 which will give + & -18

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