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nalin [4]
4 years ago
6

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

rom a railroad crossing. And wouldn’t you know it, a train moving at 30 m/s was only 60 m from the crossing. In a split second, I realized that the train was going to beat me to the crossing and that I didn’t have enough distance to stop. My only hope was to accelerate enough to cross the tracks before the train arrived. If my reaction time before starting to accelerate was 0.50 s, what minimum acceleration did my car need for me to be here today writing these words?
Physics
1 answer:
cricket20 [7]4 years ago
7 0

Answer:

a=4.44\frac{m}{s^2}

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

d=vt\\d=20\frac{m}{s}(0.5s)=10m

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}

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A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)
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a) F=(3675i-4543k)N

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k

Therefore it has 3 components:

r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2

We start by finding the velocity of the UFO, which is given by the derivative of the position:

v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t

And then, by differentiating again, we find the acceleration:

a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6

The weight of the UFO is W = 12,500 N, so its mass is:

m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg

Therefore, the components of the force on the UFO are given by Newton's second law:

F=ma

So, Substituting t = 2 s, we find:

F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N

So the net force on the UFO at t = 2 s is

F=(3675i-4543k)N

b)

The magnitude of a 3-dimensional vector is given by

|v|=\sqrt{v_x^2+v_y^2+v_z^2}

where

v_x,v_y,v_z are the three components of the vector

In this problem, the three components of the net force are:

F_x=3675 N\\F_y=0\\F_z=-4543 N

Therefore, substituting into the equation, we find the magnitude of the net force:

|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N

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