Question:
The position of a particle as a function of time t, is given by
\(x(t)=at+bt^2-ct^3\)
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
The position of a particle as a function of time t, is given by
\(x(t)=at+bt^2-ct^3\)
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
\(x(t)=at+bt^2-ct^3\)
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
Show Hint
Differentiate x twice, set the acceleration to zero to get t, and substitute that t into v(t).
Updated On: Nov 5, 2026
- \(a+\dfrac{b^2}{c}\)
- \(a+\dfrac{b^2}{4c}\)
- \(a+\dfrac{b^2}{3c}\)
- \(a+\dfrac{b^2}{2c}\)
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The Correct Option is C
Solution and Explanation
\(v=\dfrac{dx}{dt}=a+2bt-3ct^2\), \(\;\text{acceleration}=\dfrac{dv}{dt}=2b-6ct\).
Acceleration = 0 \(\Rightarrow\) \(t=\dfrac{b}{3c}\).
\(v=a+2b\cdot\dfrac{b}{3c}-3c\cdot\dfrac{b^2}{9c^2}=a+\dfrac{2b^2}{3c}-\dfrac{b^2}{3c}=a+\dfrac{b^2}{3c}\).
Answer: (3) \(a+\dfrac{b^2}{3c}\)
Acceleration = 0 \(\Rightarrow\) \(t=\dfrac{b}{3c}\).
\(v=a+2b\cdot\dfrac{b}{3c}-3c\cdot\dfrac{b^2}{9c^2}=a+\dfrac{2b^2}{3c}-\dfrac{b^2}{3c}=a+\dfrac{b^2}{3c}\).
Answer: (3) \(a+\dfrac{b^2}{3c}\)
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