Question

Utkarsh riding his bicycle with speed of 36 km/h, suddenly he saw a nail in front of his bicycle at a distance of 10m. He applies the brake to avoid the puncture in his bicycle and stops the bicycle in 2 second just before the nail find the acceleration applies through the brake

a. 10m/s^2
b. 18m/s^2
c. 3.6m/s^2
d. 5m/s^2

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Answer 1

Provided that:

• Initial velocity = 36 kmph
• Final velocity = 0 mps
• Distance = 10 metres
• Time = 2 seconds

To calculate:

• Acceleration

Solution:

• Acceleration = -5 mps sq.

Using concepts:

• Formula to convert kmph-mps:

• ${\small{\underline{\boxed{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}$

• Acceleration formula:

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

• First equation of motion:

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

• Second equation of motion:

• ${\small{\underline{\boxed{\sf{s\: = ut \: + \dfrac{1}{2} \: at^2}}}}}$

• Third equation of motion:

• ${\small{\underline{\boxed{\sf{v^2 \: - u^2 \: = 2as}}}}}$

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity, s denotes displacement or distance or height and t denotes time taken.

To calculate the acceleration we can use first equation of motion or second equation of motion or third equation of motion or acceleration formula.

• Choice may vary!

Required solution:

~ Firstly let us convert kmph-mps!

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

• Therefore, Initial velocity = 10 metre per second.

~ Now calculating acceleration!

By using acceleration formula.

→ a = (v-u)/t

→ a = (0-10)/2

→ a = -10/2

→ a = -5 m/s sq.

→ Acceleration = -5 m/s sq.

By using first equation of motion.

→ v = u + at

→ 0 = 10 + a(2)

→ 0 - 10 = 2a

→ -10 = 2a

→ -10/2 = a

→ Acceleration = -5 m/s sq.

By using second equation of motion.

→ s = ut + ½ at²

→ 10 = 10(2) + ½ × a × (2)²

→ 10 = 10(2) + ½ × 4a

→ 10 = 20 + 1 × 2a

→ 10 = 20 + 2a

→ 10 - 20 = 2a

→ -10 = 2a

→ -10/2 = a

→ Acceleration = -5 m/s sq.

By using third equation of motion.

→ v² - u² = 2as

→ (0)² - (10)² = 2a(10)

→ 0 - 100 = 20a

→ -100 = 20a

→ -10 = 2a

→ -10/2 = a

→ Acceleration = -5 m/s sq.

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• Dear users I have solve this question for you by all methods so choice may yours that which method you want to apply. Thank you!