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3 years ago

8

Plane 1 leaves city A at 7 am and flies at a speed of 250 mph in a straight line towards city B, located 1000 miles North of cit

y A. Plane 2 leaves city C located 300 miles East and 400 miles North of city A at 7:30 am and flies West at a speed v mph. The planes fly at the same altitude. For which value of v do they collide ?

1 answer:

Reika [66]3 years ago

8 0

Answer:v=142.85 mph

Explanation:

Given

Distance between two cities is 1000 miles

Speed of plane 1 id 250 mph

Plane 1 leaves at 7 am and Plane 2 leaves at 7:30 am

They will collide if plane 2 travels 300 miles and Plane 1 travels 400 miles in t and t+0.5 hr respectively

Time taken by plane 2

$t=\frac{300}{v}$ ----1

time taken by plane 1 to travel 400 miles

$t+0.5=\frac{400}{250}$ -----2

substitute value of t in 2

$\frac{300}{v}+0.5=\frac{400}{250}$

$\frac{300}{v}=2.1$

$v=\frac{300}{2.1}=142.85 mph$

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Answer with explanation:

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Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

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$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

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$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

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