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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?
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so, solving that equation for R, we get: \[\Large R = \frac{{m{v^2}}}{{2d}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now, 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: \[\Large \frac{1}{2}mv_1^2 = \frac{{m{v^2}}}{{2d}}{d_1}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0finally, after a simplification, we can write: \[\Large {d_1} = d{\left( {\frac{{{v_1}}}{v}} \right)^2}\] where: \[\Large {v_1} = 70\;Km/h\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please substitute your data into that formula and you will get the requested distance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There 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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