KCET · Maths · Application of Derivatives
The distance ' \(s\) ' in meters travelled by a particle in ' \(t\) ' seconds is given by \(s=\frac{2 t^3}{3}-18 t+\frac{5}{3}\). The acceleration when the particle comes to rest is
- A $10 \mathrm{~m}^2 / \mathrm{s}
- B \(12 \mathrm{~m}^2 / \mathrm{s}\)
- C \(18 \mathrm{~m}^2 / \mathrm{s}\)
- D \(3 \mathrm{~m}^2 / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(12 \mathrm{~m}^2 / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given,
\(\begin{aligned} & s=2 t^3-18 t+\frac{5}{3} \\ \Rightarrow \quad & \frac{d s}{d t}=6 t^2-18 \\ \Rightarrow \quad \frac{d^2 s}{d t^2} & =12 t\end{aligned}\)
We know that
\(\begin{array}{ll}\because \quad & v=u+a t \\ \Rightarrow \quad & 0=6 t^2-18+12 t^2 \\ \Rightarrow \quad 18 t^2 & =18 \\ \Rightarrow \quad & t=1 \mathrm{~s}\end{array}\)
After \(1 \mathrm{~s}\), particles comes to rest
So,
\(a=12 t\)
[From Eq. (i)]
\(\begin{aligned} & a=-12 \times 1 \mathrm{~m}^2 / \mathrm{s} \\ & a=12 \mathrm{~m}^2 / \mathrm{s}\end{aligned}\)
\(\begin{aligned} & s=2 t^3-18 t+\frac{5}{3} \\ \Rightarrow \quad & \frac{d s}{d t}=6 t^2-18 \\ \Rightarrow \quad \frac{d^2 s}{d t^2} & =12 t\end{aligned}\)
We know that
\(\begin{array}{ll}\because \quad & v=u+a t \\ \Rightarrow \quad & 0=6 t^2-18+12 t^2 \\ \Rightarrow \quad 18 t^2 & =18 \\ \Rightarrow \quad & t=1 \mathrm{~s}\end{array}\)
After \(1 \mathrm{~s}\), particles comes to rest
So,
\(a=12 t\)
[From Eq. (i)]
\(\begin{aligned} & a=-12 \times 1 \mathrm{~m}^2 / \mathrm{s} \\ & a=12 \mathrm{~m}^2 / \mathrm{s}\end{aligned}\)
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