document.write( "Question 1207344: The distance. S metres, in which a car can stop is related to its speed, 'km/h by\r
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document.write( "S=pV+ qv² where p and q are constants. A car travelling at 10 km/h can stop in a distance of 5 metres and at 20 km/h in a distance of 12 metres. Calculate the distance in which a car travelling at 30 km/h can stop \n" );
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Algebra.Com's Answer #845171 by ikleyn(52781)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( "he distance. S metres, in which a car can stop is related to its speed, 'km/h by \n" ); document.write( "S=pV+ qv² where p and q are constants. A car travelling at 10 km/h can stop in a distance of 5 metres \n" ); document.write( "and at 20 km/h in a distance of 12 metres. Calculate the distance in which a car travelling at 30 km/h can stop \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( "As you read the problem, write two equations\r\n" ); document.write( "\r\n" ); document.write( " 10*p + 10^2*q = 5 meters (for the speed V = 10 km/h)\r\n" ); document.write( "\r\n" ); document.write( " 20*p + 20^2*q = 12 meters (for the speed V = 20 km/h)\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Or, the same\r\n" ); document.write( "\r\n" ); document.write( " 10p + 100q = 5 (1)\r\n" ); document.write( "\r\n" ); document.write( " 20p + 400q = 12 (2)\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "To solve these equations, multiply equation (1) by 2 (both sides). Keep equation (2) as is\r\n" ); document.write( "\r\n" ); document.write( " 20p + 200q = 10 (1')\r\n" ); document.write( "\r\n" ); document.write( " 20p + 400q = 12 (2')\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "From equation (2'), subtract equation (1'). You will get\r\n" ); document.write( "\r\n" ); document.write( " 200q = 2\r\n" ); document.write( "\r\n" ); document.write( " q = 2/200 = 0.01.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Then from equation (1)\r\n" ); document.write( "\r\n" ); document.write( " 10p + 100*0.01 = 5\r\n" ); document.write( "\r\n" ); document.write( " 10p = 5 - 1 = 4\r\n" ); document.write( "\r\n" ); document.write( " p = 4/10 = 0.4.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Thus, now you know p= 0.4, q= 0.01.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Hence, at V= 30 km/h, the distance to stop is\r\n" ); document.write( "\r\n" ); document.write( " 30p + 30^2*q = 30*0.4 + 900*0.01 = 21,\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "ANSWER. The distance to stop at V 30 km/h is 21 meter.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Solved.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |