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12345 [234]
3 years ago
15

How far does a plane fly in 15 s while its velocity is changing from 145 m/s to 75 m/s at a uniform rate of acceleration?

Physics
2 answers:
kirza4 [7]3 years ago
6 0

Answer:

The plane travels 1650m.

Explanation:

Information Given:

  • The initial velocity of the plane v = 75 m/s
  • Duration of change in velocity t = 15s
  • The final velocity of the plane after 15 s, vf = 75 m/s

v_{f} =v_{0} +a*t\\75m/s=145m/s+a*15s\\a=(75m/s-145m/s)/15s\\a=-4.66m/s^{2}

Note that acceleration is negative, because the plane is in constant desacceleration.

To know how far does the plane fly, we use following formula:

s=v_{0} *t+1/2*a*t^{2} \\s=145m/s*15s+1/2*-4.66m/s^{2} *225s^{2} \\s=1650 m

MAVERICK [17]3 years ago
4 0

The solution would be like this for this specific problem:

Given:<span>
- The initial velocity of the plane </span><span>u = 75<span><span> m</span>/s</span><span>
</span></span>- Duration of change in velocity <span><span><span>t = 15</span>s</span><span>
</span></span>- The final velocity of the plane after 15 s, <span>v = 145<span><span> m</span>/s</span><span>
</span></span>- Let its uniform acceleration be <span>am<span>s<span>^−2</span></span></span>

<span>
Now we know that </span><span>v = u + at ⇒ .145 = 75 + a × 15</span>

<span>
</span><span>⇒ a =<span> (145) / 15</span> =<span> 14/3</span>m<span>s<span>^−2</span></span></span>

<span>
Now distance traversed during 15s is:</span>

<span>
</span><span>s = u × t +<span> 1/2</span> × a ×<span><span> t</span><span>^2</span></span> = 75 × 15 +<span><span> ½ ×<span> 14/3</span> ×<span> 15<span>^2</span></span> = 1650</span><span>m</span></span></span>

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