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lara [203]
4 years ago
7

You’re driving your car towards an intersection. A Porsche is stopped at the red light. You’re traveling at 36 km/h (10 m/s). As

you are 15 m from the light, the light turns green, and the Porsche accelerates from rest at 3 m/s2. You continue at constant speed. a. How far from the stop line do you pass the Porsche? At what time, measured from when the light turned green, do you pass the Porsche? b. As the Porsche keeps accelerating, it eventually catches up to you again. How far from the stop line does it pass you? At what time, measured from when the light turned green, does it pass you? c. If a Boston police officer happens to get you and the Porsche on a radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/h.
Physics
1 answer:
34kurt4 years ago
6 0

Your position at time t, relative to the stop line:

x_1=-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t

The Porsche's position:

x_2=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2

a. You pass the Porsche immediately after the time it takes for x_1=x_2:

-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=2.3\,\rm s

at which point you both will have traveled 7.8 m from the stop line.

b. The equation in part (a) has two solutions. The Porsche passes you at the second solution of about t=4.4\,\rm s, at which point you both will have traveled 29 m.

c. At time t, the Porsche is moving at velocity

v=\left(3\dfrac{\rm m}{\mathrm s^2}\right)t

so that at the moment it passes you, its speed is 13 m/s, which is about 46.8 km/h and below the speed limit, so neither of you will be pulled over.

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F=ma (Newton's second law of motion).

Then,
a = F/m = Uk*mg/m = Uk*g = 0.0751*9.81 = 0.737 m/s^2 (should be negative since a deceleration is expected).

v = Sqrt (u^2-2ad), Where v = final velocity, u=initial velocity, a = acceleration, d = distance moved.

Therefore,

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An object, initially at rest moves 250m in 17s. What is it's acceleration?
Gekata [30.6K]
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

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At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is
Katyanochek1 [597]

Answer:

4.47\ \text{km/h}

Explanation:

\dfrac{da}{dt} = Rate at which the distance between A and starting point of B is changing = -20 km/h

\dfrac{db}{dt} = Rate at which the distance of B is changing = 15 km/h

\dfrac{dc}{dt} = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = 20\times 4=80\ \text{km}

a = Distance remaining to the start point of B = 110-80=30\ \text{km}

b = Distance covered by B in 4 hours = 15\times 4=60\ \text{km}

Distance between A and B after 4 hours

c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}

c^2=a^2+b^2

Differentiating with respect to time we get

c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}

The rate at which the distance between the ships is changing at 4 PM is 4.47\ \text{km/h}.

7 0
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