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baherus [9]
3 years ago
11

What is the acceleration of a vehicle that changes its velocity from 100 km/h to a dead stop in 30 s ? Answer in units of m/s 2

.
Physics
1 answer:
Dafna1 [17]3 years ago
3 0

Answer:

a = -\frac{25}{27} m/s²

Explanation:

Acceleration (a) = (final velocity - initial velocity) ÷ (final time - initial time) =  \frac{v_{f} - v_{i}  }{t_{f} - t_{i} }

Converting the speed from km/h to m/s;

100 km = 100000m

1hr = 3600s

∴ 100km/hr ⇄ \frac{100000 m}{3600 s} = \frac{250}{9} m/s

Acceleration = \frac{0 - 250/9}{30} = -\frac{250}{9 * 30} = -\frac{25}{27} m/s²

i.e deceleration = \frac{25}{27} m/s²

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Answer:

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Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

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Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

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Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

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