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anonymous

  • one year ago

A car going 50km/h can stop over 15m. On the same road, in what distance can the same car be stopped when its speed is 70km/h?

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  1. Michele_Laino
    • one year ago
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    the car can stop due to the work done by the friction forces. Now the magnitude R, of such friction forces are given by the subsequent fromula: \[\Large \frac{1}{2}m{v^2} = Rd\] where m is the mass of the car and d= 15 meters, v is the velocity of the car

  2. Michele_Laino
    • one year ago
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    so, solving that equation for R, we get: \[\Large R = \frac{{m{v^2}}}{{2d}}\]

  3. Michele_Laino
    • one year ago
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    now, when the speed of the car is v_1=70 km/h, then the subsequent equation holds: \[\Large \frac{1}{2}mv_1^2 = R{d_1}\] where d_1 is the new distance, and we have to compute it. Substituting for R, we get: \[\Large \frac{1}{2}mv_1^2 = \frac{{m{v^2}}}{{2d}}{d_1}\]

  4. Michele_Laino
    • one year ago
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    finally, after a simplification, we can write: \[\Large {d_1} = d{\left( {\frac{{{v_1}}}{v}} \right)^2}\] where: \[\Large {v_1} = 70\;Km/h\]

  5. Michele_Laino
    • one year ago
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    please substitute your data into that formula and you will get the requested distance

  6. anonymous
    • one year ago
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    There is a simple formula for an object in acceleration/ retardation: v^2=u^2+2aS Where ' v=final velocity ( 0 here, as it is stopping) u=initial velocity (50kmph= (50*5/18 )m/s here as that was the initial 'speed') a= the acceleration that is acting on the vehicle (unknown, we'll find it out) S= distance over which the acceleration works (15m) Putting these values: 0^2=(50*5/18)^2+2*a*15 We get a=-6.43004 (m/s^2) This is a constant as the breaks of the car shall provide constant acceleration. For the next case, we have v=0, u=70*5/18 m/s, a=-6.4301 m/s^2, s=unknown Putting values: 0^2=(70*5/18)^2+2*-6.4301*S S=29.3997 m

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