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3 years ago

A truck decelerates at a rate of 2 m/s2. If it starts with a velocity of 15 m/s, how long will it take to stop?

1 answer:

7 0

Use the formula Vf = Vo + at

Stop => Vf = 0

=> 0 = Vo + at => Vo = -at => t =- Vo / a

Vo = 15 m/s

a = - 2 m/s^2

=> t = - 15 m/s / ( - 2 m/s^2) = 7.5 s

And you can find the distance travled by using the equation: d = Vo*t + at^2 / 2

=> d = 15 m/s * 7.5 s - 2m/s^2 * (7.5s)^2 / 2 = 56.25 m

Answer: 7.5s (and 56.25m)

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Answer:

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Explanation:

Given that

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$C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)$

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3 years ago

A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.

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Answer:

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

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$m_1u_1+m_2u_2=(m_1+m_2)V$

V is the common velocity after the collisions

$V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s$

So, the two car train will move with a common velocity of 6.79 m/s.

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Explanation:

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