Calc1231W13T3 - Distance and speed

Sunday, 21 October 2018

6:38 PM

Untitled picture.png Brandishing your mathematical skills, you find a summer job helping out with the 2019 New Years Eve fireworks display on Sydney Harbour' 503m The Sydney Harbour Bridge The lower arch of the Sydney Harbour Bridge is 503 metres in width and 118 metres high. It is also a parabola. If the zeros of the parabola are at x 0 and x 503, then the vertex is at the point [251.5,118] O _ Hence the equation of the parabola is y We will use this equation in the next few questions. Note: the Maple syntax for the point [1, 2] is (1, 2] _ Untitled picture.png Suppose a flaming ball is set to fire from [0,0] and fixed to the rail of the lower arch of the Sydney Harbour Bridge (ie the trajectory ofthe ball follows the rail). The ball must reach the point [503, 0] in exactly 6 seconds The x and y co-ordinates can be specified as a function oft in seconds, as • x(t) — . y(t) 503/6*t 118 t(t — 6). 9 dc Note: We are assuming a constant x-velocity, ie should be constant. The distance travelled by the ball at time u seconds is given by the formula Hence after 2 seconds the ball has travelled (to the nearest metre) 199 Note: you will probably want to use Maple of WolframAlpha to calculate this integral. 2 dt. O _4_ metres_ Untitled picture.png 503 6 118 9 int( sqrt(diflx round ( %) -6) -6) 472 503 6 118 9 difly: t)2) 475793 4 253009 ln(503) — 944 253009 475793 ) 36 199 2499865 253009 ln(3) — 253009 ln(472 2499865 ) Untitled picture.png 503 6 118 9 int( sqrt(diflx round ( %) -6) -6) 472 503 6 118 9 difly: t)2) 475793 4 253009 ln(503) — 944 253009 475793 ) 36 199 2499865 253009 ln(3) — 253009 ln(472 2499865 )
Untitled picture.png Suppose a flaming ball is set to fire from [0,0] and fixed to the rail of the lower arch of the Sydney Harbour Bridge (ie the trajectory ofthe ball follows the rail). The ball must reach the point [503, 0] in exactly 6 seconds The x and y co-ordinates can be specified as a function oft in seconds, as • x(t) — . y(t) 503/6*t 118 t(t — 6). 9 dc Note: We are assuming a constant x-velocity, ie should be constant. The distance travelled by the ball at time u seconds is given by the formula Hence after 2 seconds the ball has travelled (to the nearest metre) 199 Note: you will probably want to use Maple of WolframAlpha to calculate this integral. 2 dt. O _4_ metres_ Untitled picture.png The speed of the flaming ball is the rate of change in distance travelled_ Hence the speed of the flare at time u seconds is: d • After 2 seconds the ball is moving ata speed of 88 O metres per second (to the nearest integer). • The initial speed of the ball is 115 O metres per second (to the nearest integer) Untitled picture.png 472 Recall: that the equation describing the lower arch of the bridge is y 253009 The average height above launch level of the ball is (to the nearest metre) 79 + 118. O metres. Untitled picture.png Recall the formula for the speed at time u seconds is given by: du To accommodate even more fireworks in the 2019 schedule, the artistic director requires that the ball must now take only 3 seconds to exactly follow the trajectory of the lower arch. Hence the new initial speed is (to the nearest integer): 230 O metres per second Untitled picture.png 503 6 118 9 int( sqrt(diflx round ( %) -6) -6) 472 503 6 118 9 difly: t)2) 475793 4 253009 ln(503) — 944 253009 475793 ) 36 199 2499865 253009 ln(3) — 253009 ln(472 2499865 ) Untitled picture.png = 2, sqrt(djfis = O, sqrt(dtfis diff y, t)2))) 88 diff y, t)2))) 115 Untitled picture.png М -472/253009 503 round( %) 78_66666666 79 Untitled picture.png 503 3 subs(x = O, %) 503 944 x + 0.9383697813 253009 229.9258334 Untitled picture.png 503 6 118 9 int( sqrt(diflx round ( %) -6) -6) 472 503 6 118 9 difly: t)2) 475793 4 253009 ln(503) — 944 253009 475793 ) 36 199 2499865 253009 ln(3) — 253009 ln(472 2499865 ) Untitled picture.png 503 3 subs(x = O, %) 503 944 x + 0.9383697813 253009 229.9258334

 

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