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hodyreva [135]
3 years ago
13

a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?

Physics
1 answer:
astra-53 [7]3 years ago
3 0
We are given the following conditions:
v_{0} = 28m/s

t = 4s

v_{f} = 0m/s

Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.

So we look at our kinematics equation that contains the given variables.

v_f = v_0 + at

This contains our target variable and our initial conditions. Plug and chug, we get:

(0m/s) = (28m/s) + a(4s) -28m/s = (4s)*a a = -7m/s^2

Thus the answer is:

a = -7m/s^2

Hope this helps!


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Answer:

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Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

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∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

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<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

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<em>f</em> = <em>µn</em>

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<em>a</em> = 5.80 m/s²

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