Answer:
no:
Explanation:
it would grow and no longer be able to fit through the loop due to the hot air expanding.
Till the time car is just adjacent to the bicycle we can say
distance moved by cycle = distance moved by car
Time taken by car to accelerate from rest


Time taken by cycle to accelerate

now the distance moved by cycle in time "t"

distance moved by car in same time

now make them equal



so cycle will move ahead of car for t = 5.68 s
Answer:
The last two bearings are
49.50° and 104.02°
Explanation:
Applying the Law of cosine (refer to the figure attached):
we have
x² = y² + z² - 2yz × cosX
here,
x, y and z represents the lengths of sides opposite to the angels X,Y and Z.
Thus we have,

or

substituting the values in the equation we get,

or

or
X = 26.47°
similarly,

or

or
Y = 49.50°
Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°
The bearing of 2 last legs of race are angels Y and Z.