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3 years ago

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Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

at the +x-axis is directed eastward and the +y-axis is directed northward. Find the magnitude of the velocity vector of the truck relative to the SUV.

1 answer:

mario62 [17]3 years ago

5 0

Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

Velocity of truck $v_t=16\hat{i}$

SUV is moving south with a velocity of 20 m/s

Velocity of SUV in vector form $v_s=-20\hat{j}$

Velocity of truck relative to the SUV

$v_{ts}=v_{t}-v_s$

$v_{ts}=16\hat{i}-(-20\hat{j})$

Magnitude of relative velocity is

$|v_{ts}|=\sqrt{16^2+20^2}$

$|v_{ts}|=25.61\ m/s$

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$g = \frac{GM}{(d-R_{CM})^2}$

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At the same time we have that centripetal acceleration is given as

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ivann1987 [24]

Answer:

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Explanation:

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speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

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now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

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Given :

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