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neonofarm [45]
3 years ago
7

Two cars collide on the parkway. The drivers are following too closely and the driver in the front slows because of traffic to a

velocity of 20.0m/s and the driver behind is still moving at 30.0m/s, both cars traveling in the same direction in a rear end collision. If the cars are stuck together after the collision, how fast will they be traveling?
Physics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

The are moving with a speed of 25 m/s after the collision.

Explanation:

assuming the mass (m) of the cars are equal, let v = 20 m/s be the velocity of the driver in the front and u = 30 m/s be the velocity of the driver behind. let V be the velocity of the cars after the collision.

then, by conservation of linear momentum:

m×(v + u) = V×2×m

(v + u) = V×2

V = (v + u)/2

  = (20 + 30)/2

  = 25 m/s

Therefore, the are moving with a speed of 25 m/s after the collision.

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slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

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3 years ago
Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the wa
rewona [7]

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

v = \dfrac{d}{10.8-t}

d = v(10.8-t)

The distance is 549 and maximum speed is 65 mi/h.

549 = 65(10.8 - t)

10.8-t = 8.45

t = 10.8 - 8.45 = 1.55 \text{ h}

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