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Anuta_ua [19.1K]

3 years ago

14

A car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s2 to reach a speed of 12.0 m/s. how long does it take this accelerati

on to occur

1 answer:

lina2011 [118]3 years ago

4 0

Vf=Vi+at
12=7+2.5t
5=2.5t
t=2s

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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

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<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 = ?

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<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>

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

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➪ $\sf 2.25 - 0 = 4 \times a$

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➪ $\sf 2.25 = 4 \times a$

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➪ $\sf a = \dfrac{2.25}{4}$

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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

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