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2 years ago

7

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.60 m/s. The car is a distan

ce d away. The bear is 39.3 m behind the tourist and running at 5.00 m/s. The tourist reaches the car safely. What is the maximum possible value for d

1 answer:

In-s [12.5K]2 years ago

7 0

Answer:

101 meters

Explanation:

Distance traveled by the tourist:

d = 3.60 m/s × t

Distance traveled by the bear:

d + 39.3 m = 5.00 m/s × t

Substitute:

3.6 t + 39.3 = 5 t

39.3 = 1.4 t

t = 28.1

d = 101

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Answer with explanation:

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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

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

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