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3 years ago

9

Formula one racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.

a formula one racer traveling at 90 m/s can stop in a distance of 110 m. What is the magnitude of the car's acceleration as it slows during braking?

1 answer:

soldi70 [24.7K]3 years ago

8 0

Answer:

a = -36.8 m/s/s

Explanation:

Initial speed of the car

$v_i = 90 m/s$

finally car will stop after it cover the distance

$d = 110 m$

so we have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - 90^2 = 2(a) (110)$

$a = \frac{90^2}{220}$

$a = -36.8 m/s^2$

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Answer:

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Given:

$\lambda = 6.4 \times 10^{-7} m$

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To find:

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Formula used:

y = $\frac{m \times \lambda \times D}{d}$

y = distance of first bright band from central maxima

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Thus,

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d = $\frac{1 \times 6.4 \times 10^{-7} \times 4 }{2 \times 10^{-2} }$

d = 1.28 × $10^{-4}$

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An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel

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I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door $d$ )...

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We also have

$\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd$

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$d$ is fixed, so

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At the point when $\theta=40^\circ$ and $\dfrac{\mathrm db}{\mathrm dt}=1.8$ ft/s, we get

$\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}$

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