VW Beetle beats Porsche 911 in 1 mile race! In an unprecedented race conducted by the renowned Top Gear team, a VW Beetle (circa 1967) gets the jump on a Porsche 911 Turbo. To even up the odds, while the Porsche is driven along a standard mile-long desert track, the Beetle gets a helping hand from gravity by being dropped by helicopter, 1 mile above the finish line. The Porsche should be able to complete a drag race such as this in around 37 seconds. Assuming air resistance is negligible, calculate the time taken for the Beetle to reach the finish line. Use 1 ππππ β 1600 πππ‘πππ . Using energy considerations, calculate the maximum speed of the Beetle. In practice, the top speed of the 840kg Beetle was 49.7ππ β1 (roughly 110mph). Calculate the work done against resistive forces during the motion. A better model of the situation takes into account air resistance, which is given by: π = ππ£ 2 , where π is the force of air resistance and π£ is the velocity. After 8 seconds, the Beetle is experiencing an acceleration of 1.46ππ β2 and is travelling at a speed of 45ππ β2 . Show that π = 3.46 π‘π 3 π . π. It can be shown that the velocity of a body of mass π, dropped from rest and falling under gravity, which experiences a resistance force π = ππ£ 2 is given by: π£= ππ tanh π ππ π‘ π Use the standard result tanh π₯ ππ₯ = ln cosh π₯ to find an expression for the displacement of such a body at time π‘, and hence calculate the time taken for the Beetle to reach the ground. VW Beetle beats Porsche 911 in 1 mile race! SOLUTIONS In an unprecedented race conducted by the renowned Top Gear team, a VW Beetle (circa 1967) gets the jump on a Porsche 911 Turbo. To even up the odds, while the Porsche is driven along a standard mile-long desert track, the Beetle gets a helping hand from gravity by being dropped by helicopter, 1 mile above the finish line. The Porsche should be able to complete a drag race such as this in around 37 seconds. Assuming air resistance is negligible, calculate the time taken for the Beetle to reach the finish line. Use 1 ππππ β 1600 πππ‘πππ . π = 1600 π’ = 0 π£ = β π = 9.8 π‘ =? 1 π = π’π‘ + ππ‘ 2 2 βΉ 1600 = 4.9π‘ 2 βΉ π = ππ. ππ ππ π π. π. Using energy considerations, calculate the maximum speed of the Beetle. π΄ππ πΊππΈ ππ ππππ£πππ‘ππ π‘π πΎπΈ: 1 πππ = ππ£ 2 2 βΉ π£ = 2 × 1600 × 9.8 = πππππβπ β 400πππ In practice, the top speed of the 840kg Beetle was 49.7ππ β1 (roughly 110mph). Calculate the work done against resistive forces during the motion. πΌπππ‘πππ ππππππ¦ = πΊππΈ = 9.8 × 1600 × 840 = 13,171,200π½ 1 πΉππππ ππππππ¦ = πΎπΈ = 49.72 × 840 = 1,037,437.8π½ 2 πΏππ π‘ π‘π πππ ππ π‘ππ£π ππππππ = 13,171,200 β 1,037,437.8 = ππ, πππ, πππ, ππ± A better model of the situation takes into account air resistance, which is given by: π = ππ£ 2 , where π is the force of air resistance and π£ is the velocity. After 8 seconds, the Beetle is experiencing an acceleration of 1.46ππ β2 and is travelling at a speed of 45ππ β2 . Show that π = 3.46 π‘π 3 π . π. π ππ π’ππ‘πππ‘ πππππ = 840π β ππ£ 2 840π = 840π β ππ£ π = 1.46 π€πππ π£ = 45 βΉ 2 ππ£ 2 π=πβ 840 βΉ π 452 1.46 = π β 840 βΉ = 3.46 π‘π 3 π . π. It can be shown that the velocity of a body of mass π, dropped from rest and falling under gravity, which experiences a resistance force π = ππ£ 2 is given by: π£= ππ tanh π ππ π‘ π Use the standard result tanh π₯ ππ₯ = ln cosh π₯ to find an expression for the displacement of such a body at time π‘, and hence calculate the time taken for the Beetle to reach the ground. π₯= βΉ π₯= π£ ππ‘ = ππ π ππ tanh π ππ π‘ π π ππ ln cosh π‘ ππ π π ππ π₯ = ln cosh π‘ π π πΉππ π‘ππ π΅πππ‘ππ: π‘ = ππ‘ = ππ π tanh ππ π‘ π ππ‘ + πΆ (π ππππ π₯ = 0 ππ‘ π‘ = 0, πΆ = 0) βΉ π‘= ππ₯ π β1 π cosh π ππ 3.46×1600 840 β1 cosh π 840 = ππ. ππ ππ π π. π. 3.46 × 9.8
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