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nalin [4]
3 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]3 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 parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr
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Q stay the same

Explanation:

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In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis
IRINA_888 [86]

Answer:

Compression Test

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3 years ago
Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys
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Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

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Find the distance covered by Jenny, when Alyssa starts

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Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

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3 years ago
Read 2 more answers
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