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Vlad1618 [11]
3 years ago
8

A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 seconds later?

Physics
1 answer:
Bogdan [553]3 years ago
4 0
Given: 

V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
V2 = final velocity = ?

For linear-motion problems with those given terms, the following formula is used:

V2 = V1 + as

Substituting the given values:

V2 = 24 + 2(8)
V2 = 24 + 16
V2 = 40 m/s

Therefore the car will have a speed of 40 m/s as it leaves the tunnel.
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1)

Answer:

Part 1)

H = 30.6 m

Part 2)

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Part 3)

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initial speed of the ball upwards

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H = \frac{24.5^2}{2(9.80)}

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Part 2)

As we know that final speed will be zero at maximum height

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Part 3)

Since the time of ascent of ball is same as time of decent of the ball

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a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

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Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

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final speed of the rocket is given as

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Part 1)

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Part 2)

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Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

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y = \frac{1}{2}gt^2

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y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

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