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kifflom [539]
3 years ago
15

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

and slows at 7.0 m/s2. How far does the car go before it stops?
Physics
1 answer:
mr Goodwill [35]3 years ago
3 0

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

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\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

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\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

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\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

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\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

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<u>Let's Slove :</u><u> </u>

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  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

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  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
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