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Free_Kalibri [48]
2 years ago
10

The wheelchair starts from rest. It accelerates at a constant rate until it has a speed of 1.5 m/s. The wheelchair travels a dis

tance of 2.0 m while it is accelerating. Calculate the acceleration of the wheelchair.
Please help me i have a test and i have zero idea how to do thiss............
Physics
1 answer:
anastassius [24]2 years ago
7 0
<h2><em><u>Answ</u></em><em><u>er</u></em><em><u>:</u></em><em><u>-</u></em></h2>

\pink{\bigstar} The acceleration of the wheelchair is \large\leadsto\boxed{\tt\purple{0.5625 \: m/s^2}}

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• <u>Given</u><u>:</u><u>-</u></h3>

  • Initial velocity of the wheelchair = 0 m/s

  • Final velocity of the wheelchair = 1.5 m/s

  • Distance covered by the wheelchair = 2 m

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• <u>To</u><u> </u><u>Find</u><u>:</u><u>-</u></h3>

  • Acceleration of the wheelchair = ?

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• <u>Solution</u><u>:</u><u>-</u></h3>

We know,

• <u>Third</u><u> </u><u>Eq</u><u>uation</u><u> of</u><u> Motion</u><u>:</u><u>-</u>

\pink{\bigstar} \large\underline{\boxed{\bf\green{v^2 - u^2 = 2as}}}

where,

  • u = Initial velocity

  • v = Final velocity

  • a = Acceleration

  • s = Distance covered

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• <u>S</u><u>u</u><u>b</u><u>s</u><u>t</u><u>i</u><u>t</u><u>u</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>the</u><u> </u><u>values</u><u> </u><u>in</u><u> </u><u>th</u><u>e</u><u> </u><u>Formula</u><u>:</u><u>-</u>

➪ \sf (1.5)^2 - (0)^2 = 2 \times a \times 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ \sf 2.25 - 0 = 4 \times a

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ \sf 2.25 = 4 \times a

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ \sf a = \dfrac{2.25}{4}

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ \large{\bold\red{a = 0.5625 \: m/s^2}}

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the acceleration of the wheelchair is 0.5625 m/s².

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Explanation:

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⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

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Applying logarithm on both sides,

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⇒log(2)=\frac{2}{\frac{L}{R}}

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Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

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⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

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also log4=log2^{2}=2log2

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3 years ago
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Answer:

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For the above value of pressure ratio

Obtain the area ratio from the isentropic flow table

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The value of pressure ratio is Ae/A* = 1.115

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Answer:

<h2> 0.147136N/m²</h2>

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