VW Beetle beats Porsche 911 in 1 mile race!

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 π‘šπ‘–π‘™π‘’ β‰ˆ 1600 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘ .
Using energy considerations, calculate the maximum speed of the Beetle.
In practice, the top speed of the 840kg Beetle was 49.7π‘šπ‘  βˆ’1 (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: 𝑅 = π‘˜π‘£ 2 , where 𝑅 is the force of air resistance and 𝑣 is the velocity.
After 8 seconds, the Beetle is experiencing an acceleration of 1.46π‘šπ‘  βˆ’2 and is
travelling at a speed of 45π‘šπ‘  βˆ’2 . Show that π‘˜ = 3.46 π‘‘π‘œ 3 𝑠. 𝑓.
It can be shown that the velocity of a body of mass π‘š, dropped from rest and
falling under gravity, which experiences a resistance force 𝑅 = π‘˜π‘£ 2 is given by:
𝑣=
π‘šπ‘”
tanh
π‘˜
π‘˜π‘”
𝑑
π‘š
Use the standard result tanh π‘₯ 𝑑π‘₯ = ln cosh π‘₯ to find an expression for the
displacement of such a body at time 𝑑, 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 π‘šπ‘–π‘™π‘’ β‰ˆ 1600 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘ .
𝑠 = 1600 𝑒 = 0 𝑣 = βˆ’ π‘Ž = 9.8 𝑑 =?
1
𝑠 = 𝑒𝑑 + π‘Žπ‘‘ 2
2
⟹
1600 = 4.9𝑑 2
⟹ 𝒕 = πŸπŸ–. πŸπ’” 𝒕𝒐 πŸ‘ 𝒔. 𝒇.
Using energy considerations, calculate the maximum speed of the Beetle.
𝐴𝑙𝑙 𝐺𝑃𝐸 𝑖𝑠 π‘π‘œπ‘›π‘£π‘’π‘Ÿπ‘‘π‘’π‘‘ π‘‘π‘œ 𝐾𝐸:
1
π‘šπ‘”π‘• = π‘šπ‘£ 2
2
⟹
𝑣 = 2 × 1600 × 9.8 = πŸπŸ•πŸ•π’Žπ’”βˆ’πŸ β‰ˆ 400π‘šπ‘π‘•
In practice, the top speed of the 840kg Beetle was 49.7π‘šπ‘  βˆ’1 (roughly 110mph).
Calculate the work done against resistive forces during the motion.
πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ = 𝐺𝑃𝐸 = 9.8 × 1600 × 840 = 13,171,200𝐽
1
πΉπ‘–π‘›π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ = 𝐾𝐸 = 49.72 × 840 = 1,037,437.8𝐽
2
πΏπ‘œπ‘ π‘‘ π‘‘π‘œ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘–π‘£π‘’ π‘“π‘œπ‘Ÿπ‘π‘’π‘  = 13,171,200 βˆ’ 1,037,437.8 = 𝟏𝟐, πŸπŸ‘πŸ‘, πŸ•πŸ”πŸ, πŸπ‘±
A better model of the situation takes into account air resistance, which is given
by: 𝑅 = π‘˜π‘£ 2 , where 𝑅 is the force of air resistance and 𝑣 is the velocity.
After 8 seconds, the Beetle is experiencing an acceleration of 1.46π‘šπ‘  βˆ’2 and is
travelling at a speed of 45π‘šπ‘  βˆ’2 . Show that π‘˜ = 3.46 π‘‘π‘œ 3 𝑠. 𝑓.
π‘…π‘’π‘ π‘’π‘™π‘‘π‘Žπ‘›π‘‘ π‘“π‘œπ‘Ÿπ‘π‘’ = 840𝑔 βˆ’ π‘˜π‘£ 2
840π‘Ž = 840𝑔 βˆ’ π‘˜π‘£
π‘Ž = 1.46 𝑀𝑕𝑒𝑛 𝑣 = 45
⟹
2
π‘˜π‘£ 2
π‘Ž=π‘”βˆ’
840
⟹
π‘˜ 452
1.46 = 𝑔 βˆ’
840
⟹ = 3.46 π‘‘π‘œ 3 𝑠. 𝑓.
It can be shown that the velocity of a body of mass π‘š, dropped from rest and
falling under gravity, which experiences a resistance force 𝑅 = π‘˜π‘£ 2 is given by:
𝑣=
π‘šπ‘”
tanh
π‘˜
π‘˜π‘”
𝑑
π‘š
Use the standard result tanh π‘₯ 𝑑π‘₯ = ln cosh π‘₯ to find an expression for the
displacement of such a body at time 𝑑, and hence calculate the time taken for the
Beetle to reach the ground.
π‘₯=
⟹
π‘₯=
𝑣 𝑑𝑑 =
π‘šπ‘”
π‘˜
π‘šπ‘”
tanh
π‘˜
π‘˜π‘”
𝑑
π‘š
π‘š
π‘˜π‘”
ln cosh
𝑑
π‘˜π‘”
π‘š
π‘š
π‘˜π‘”
π‘₯ = ln cosh
𝑑
π‘˜
π‘š
πΉπ‘œπ‘Ÿ 𝑑𝑕𝑒 𝐡𝑒𝑒𝑑𝑙𝑒: 𝑑 =
𝑑𝑑 =
π‘šπ‘”
π‘˜
tanh
π‘˜π‘”
𝑑
π‘š
𝑑𝑑
+ 𝐢 (𝑠𝑖𝑛𝑐𝑒 π‘₯ = 0 π‘Žπ‘‘ 𝑑 = 0, 𝐢 = 0)
⟹
𝑑=
π‘˜π‘₯
π‘š
βˆ’1 π‘š
cosh 𝑒
π‘˜π‘”
3.46×1600
840
βˆ’1
cosh 𝑒 840
= πŸ‘πŸ”. πŸ‘π’” 𝒕𝒐 πŸ‘ 𝒔. 𝒇.
3.46 × 9.8