Answer:-
The acceleration of the wheelchair is
![\large\leadsto\boxed{\tt\purple{0.5625 \: m/s^2}}](https://img.qammunity.org/2022/formulas/physics/high-school/w47292p9hrxncbbbiglihy3vifvv8fggdz.png)
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• Given:-
- Initial velocity of the wheelchair = 0 m/s
- Final velocity of the wheelchair = 1.5 m/s
- Distance covered by the wheelchair = 2 m
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• To Find:-
- Acceleration of the wheelchair = ?
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• Solution:-
We know,
• Third Equation of Motion:-
![\large\underline{\boxed{\bf\green{v^2 - u^2 = 2as}}}](https://img.qammunity.org/2022/formulas/physics/high-school/h33uivongfd2gabpko86pynisvs32zepsm.png)
where,
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• Substituting the values in the Formula:-
➪
![\sf (1.5)^2 - (0)^2 = 2 * a * 2](https://img.qammunity.org/2022/formulas/physics/high-school/33wcu1czxb99uwrwqd6wn1gsm3sl3pikcs.png)
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➪
![\sf 2.25 - 0 = 4 * a](https://img.qammunity.org/2022/formulas/physics/high-school/52dwctk1z1n9eft5ygprzjvmhmoh3ozh3t.png)
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➪
![\sf 2.25 = 4 * a](https://img.qammunity.org/2022/formulas/physics/high-school/5b0agclaebe42aq2ztd4da014qv79gsnpi.png)
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➪
![\sf a = (2.25)/(4)](https://img.qammunity.org/2022/formulas/physics/high-school/mp00h5cwzucl6z2o5u0jukyhuml5slajsf.png)
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★
![\large{\bold\red{a = 0.5625 \: m/s^2}}](https://img.qammunity.org/2022/formulas/physics/high-school/dtamejwsc68sstornoatsu71nnixcju3aa.png)
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Therefore, the acceleration of the wheelchair is 0.5625 m/s².