MHT CET · Maths · Differentiation
The equation of motion of a particle moving along a straight line is \(s=2 t^{3}-9 t^{2}+12 t\), where the units of \(s\) and \(t\) are centimetre and second. The acceleration of the particle will be zero after
- A \(\frac{3}{2} \mathrm{~s}\)
- B \(\frac{2}{3} \mathrm{~s}\)
- C \(\frac{1}{2} \mathrm{~s}\)
- D \(1 \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{2} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(\frac{d s}{d t}=6 t^{2}-18 t+12\)
Again, \(\frac{d^{2} s}{d t^{2}}=12 t-18=\) acceleration
If acceleration becomes zero, then
\(0 =12 t-18\)
\(\Rightarrow t =\frac{3}{2} \mathrm{~s}
\)
Hence, acceleration will be zero after \(\frac{3}{2} \mathrm{~s}\)
Again, \(\frac{d^{2} s}{d t^{2}}=12 t-18=\) acceleration
If acceleration becomes zero, then
\(0 =12 t-18\)
\(\Rightarrow t =\frac{3}{2} \mathrm{~s}
\)
Hence, acceleration will be zero after \(\frac{3}{2} \mathrm{~s}\)
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