Question

A motor cycle along with the rider weighs 2 KN, the C.G. of the machine and
rider combined being 60 cm above the ground, with the machine in vertical
position. The M.I. of each road wheel is 1030 N/mm², and the rolling diameter is
60 cm. The engine rotates at 6 times of the road wheels and in the same sense. The
M.I. of rotating parts of the engine is 165 N/mm². Determine the angle of heel
necessary if the unit is speeding at 62.5 km/h round a curve of 30.4 m.
​

Answer 1

GIVEN:

Initial Velocity (u) of an object = 20m/s

Final velocity (v) = 0

g = -10 m ( Since the object thrown vertically upward )

TO FIND:

What is the distance travelled by the object to reach the highest point. ?

What is the time taken by the object to reach the highest point ?

SOLUTION:

❶ We have to find the distance covered by the object to reach the highest point

According to question:

On putting the given values in the equation, we get

$\bf{ v^2 - u^2 = 2gh}$

$\rm{\rightharpoonup (0)^2 - (20)^2 = 2\times -10 \times h}$

$\rm{\rightharpoonup 0 - 400 = -20h}$

$\rm{\rightharpoonup \cancel{-} 400 = \cancel{-}20h}$

$\rm{\rightharpoonup h = \cancel\dfrac{400}{20}}$

$\large\bf{\rightharpoonup \star \: h = 20 \: m \: \star}$

Distance = h = 20 m

❷ Now, we have to find the time taken by the object to cover the distance

According to question:-

On putting the given values in the formula, we get

$\bf{ v = u + gt}$

$\rm{\rightharpoonup 0 = 20 + (-10) \times t}$

$\rm{\rightharpoonup 0 = 20-10 t}$

$\rm{\rightharpoonup 10t = 20}$

$\rm{\rightharpoonup t = \cancel\dfrac{20}{10}}$

$\large\bf{\rightharpoonup \star \: t = 2 \: seconds \: \star}$

Time = t = 2 seconds

❝ Hence, the distance covered by the object to reach the highest point is 20 m and the time taken by the object to reach the highest point is 2 seconds ❞

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