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3 years ago

A 90-meter train travels at a constant speed of 10 meters per second. How long will it take to cross a 0.06 bridge?

Physics

1 answer:

salantis [7]3 years ago

5 0

Answer:

The time taken for the train to cross the bridge is 9.01 s

Explanation:

Given;

length of the train, L₁ = 90 m

length of the bridge, L₂ = 0.06 m

speed of the train, v = 10 m/s

Total distance to be traveled, = L₁ + L₂

= 90 m + 0.06 m

= 90.06 m

Time of motion = Distance / speed

Time of motion = 90.06 / 10

Time of motion = 9.006 s ≅ 9.01 s

Therefore, the time taken for the train to cross the bridge is 9.01 s

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and its velocity is

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When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

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(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

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${v_f}^2-{v_i}^2=2a\Delta y$

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$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Answer:

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