Question:
A body starts from origin and moves along x-axis with velocity \(v = (4t^3 - 2t)\). What is the acceleration at distance 2 m from origin?
A body starts from origin and moves along x-axis with velocity \(v = (4t^3 - 2t)\). What is the acceleration at distance 2 m from origin?
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When velocity is given in \(t\), first find \(x(t)\), then substitute to get required \(t\).
Updated On: Apr 23, 2026
- \(28\,m/s^2\)
- \(22\,m/s^2\)
- \(12\,m/s^2\)
- \(10\,m/s^2\)
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The Correct Option is B
Solution and Explanation
Step 1: Find position.
\[ v = \frac{dx}{dt} = 4t^3 - 2t \] \[ x = \int (4t^3 - 2t)dt = t^4 - t^2 \]
Step 2: Use \(x=2\).
\[ t^4 - t^2 = 2 \] Let \(y = t^2\): \[ y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1)=0 \Rightarrow y=2 \] \[ t = \sqrt{2} \]
Step 3: Find acceleration.
\[ a = \frac{dv}{dt} = 12t^2 - 2 \] \[ = 12(2) - 2 = 24 - 2 = 22 \] Conclusion: \[ {22\,m/s^2} \]
\[ v = \frac{dx}{dt} = 4t^3 - 2t \] \[ x = \int (4t^3 - 2t)dt = t^4 - t^2 \]
Step 2: Use \(x=2\).
\[ t^4 - t^2 = 2 \] Let \(y = t^2\): \[ y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1)=0 \Rightarrow y=2 \] \[ t = \sqrt{2} \]
Step 3: Find acceleration.
\[ a = \frac{dv}{dt} = 12t^2 - 2 \] \[ = 12(2) - 2 = 24 - 2 = 22 \] Conclusion: \[ {22\,m/s^2} \]
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