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bija089 [108]
3 years ago
12

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

the tires and the level road is 0 80 how long will the skid marks be? A) 37 m B) 30 m C) 34 m D 46 m
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

v^2 -u^2 = 2ad (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

F=ma=-\mu mg

where

m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

And we can substitute it into eq.(1) to find d:

v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m

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a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
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-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
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=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
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Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
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In the frame of measurements from the boat, the boat itself must start
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The mass of the boat is 300 kg so
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Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
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The boat is moving 0.62 m/s away from the jump-off point, and away from
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To somebody standing on the dock, the whole boat, with its intrepid passenger
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same answer that I got earlier, in the frame of reference tied to the dock.

  yay !

By the way ... thanks for the 6 points.  The warm cloudy water
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