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Mice21 [21]
4 years ago
9

How is the position vs. time graph with the cart moving towards the sensor with decreasing speed different than the position vs.

time graph with the cart moving away from the sensor at increasing speed? How are the graphs similar?
Physics
1 answer:
IrinaVladis [17]4 years ago
6 0

Answer:

if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.

Y (position)

I

sensor-------------------------------------------------------------------

I                                                    o

I                                     o

I                           o

I                   o

I            o

I       o

I   o

I------------------------------------------------------------------------------------- X (time)

in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.

Y (position)

I

sensor ----------------------

I       o

I                 o

I                          o

I                                 o

I                                       o

I                                            o

I                                              o

I------------------------------------------------------------------------------------- X (time)

i hope you can understand it, kinda hard to do graphs here.

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An SUV has a mass of 6000 kg and a compact car has a mass of 2000 kg. They both go around the same unbanked circular curve on th
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Answer:

Therefore, the centripetal acceleration of the SUV is <u>3 times</u> ao of the compact car.

Explanation:

FOR SUV:

ac_{suv} = \frac{mv^2}{r}\\

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ac_{suv} = \frac{6000\ vo^2}{r}\\---------------- equation (1)

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ao = centripetal acceleration of the car = ?

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v = speed of car = vo

r = radius of path

Therefore,

ao = \frac{2000\ vo^2}{r}\\--------------------- equation (2)

Dividing equation (1) by eq(2):

\frac{ac_{suv}}{ao} = \frac{6000}{2000}\\\\ac_{surv} = 3ao

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