Question:
An object moving along \(x\)-axis with a uniform acceleration has velocity
\[
\vec{V}=(12\;\text{cm s}^{-1})\hat{i}
\]
at \(x=3\;\text{cm}\). After \(2\;\text{s}\), if it is at \(x=-5\;\text{cm}\), then its acceleration is
An object moving along \(x\)-axis with a uniform acceleration has velocity
\[
\vec{V}=(12\;\text{cm s}^{-1})\hat{i}
\]
at \(x=3\;\text{cm}\). After \(2\;\text{s}\), if it is at \(x=-5\;\text{cm}\), then its acceleration is
Show Hint
For motion with constant acceleration, always use the kinematic equation:
\[
x=x_0+ut+\frac12 at^2
\]
when position, velocity, time, and acceleration are related.
Updated On: Jun 22, 2026
- \(\vec{a}=(-16\;\text{cm s}^{-2})\hat{i}\)
- \(\vec{a}=(11\;\text{cm s}^{-2})\hat{i}\)
- \(\vec{a}=(-11\;\text{cm s}^{-2})\hat{i}\)
- \(\vec{a}=(8\;\text{cm s}^{-2})\hat{i}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the given data.
Initial position: \[ x_0=3\;\text{cm} \] Initial velocity: \[ u=12\;\text{cm s}^{-1} \] Final position after \(t=2\;\text{s}\): \[ x=-5\;\text{cm} \] Acceleration \(a\) is to be found.
Step 2: Use the equation of motion.
For uniform acceleration, \[ x=x_0+ut+\frac12 at^2 \] Substituting the given values, \[ -5=3+(12)(2)+\frac12 a(2)^2 \] \[ -5=3+24+2a \] \[ -5=27+2a \]
Step 3: Solve for acceleration.
\[ 2a=-5-27 \] \[ 2a=-32 \] \[ a=-16\;\text{cm s}^{-2} \] Hence, \[ \vec{a}=(-16\;\text{cm s}^{-2})\hat{i} \]
Step 4: Final conclusion.
Therefore, the acceleration of the object is \[ \boxed{\vec{a}=(-16\;\text{cm s}^{-2})\hat{i}} \]
Initial position: \[ x_0=3\;\text{cm} \] Initial velocity: \[ u=12\;\text{cm s}^{-1} \] Final position after \(t=2\;\text{s}\): \[ x=-5\;\text{cm} \] Acceleration \(a\) is to be found.
Step 2: Use the equation of motion.
For uniform acceleration, \[ x=x_0+ut+\frac12 at^2 \] Substituting the given values, \[ -5=3+(12)(2)+\frac12 a(2)^2 \] \[ -5=3+24+2a \] \[ -5=27+2a \]
Step 3: Solve for acceleration.
\[ 2a=-5-27 \] \[ 2a=-32 \] \[ a=-16\;\text{cm s}^{-2} \] Hence, \[ \vec{a}=(-16\;\text{cm s}^{-2})\hat{i} \]
Step 4: Final conclusion.
Therefore, the acceleration of the object is \[ \boxed{\vec{a}=(-16\;\text{cm s}^{-2})\hat{i}} \]
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