Question

Q1 A bus starting from rest moves a uniform acceleration of 0.1m-s for 2 minutes.Find(a) the speed acquired,(b)the distance travelled.​

Answer 1

Answer:

Hey Mate. ....

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here is your answer...

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$a) \: given.. \\ u = 0 \\ a = 0.1 \\ t = 2 \times 60 = 120 \\ .. \\ \implies \: a = \frac{v - u}{t} \\ .. \\ \implies \: 0.1 = \frac{v - 0}{120} \\ .. \\ \implies \: v = 120 \times 0.1 \\.. \\ \implies \: \red{ v = 12 \: m} \\ .. \\ b) \: given.. \\ v = 12 \\ a = 0.1 \\ u = 0 \\ t = 120.. \\ .. \\ so.. \\ {v}^{2} - {u}^{2} = 2as \\ .. \\ \implies \: 12 \times 12 - 0 = 2 \times 0.1 \times s \\ .. \\ \implies \: 144 = 0.2s \\ .. \\ \implies \: s = \frac{144}{0.2} \\ .. \\ \implies \: \red{s = 720 \: m}$

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Answer 2

Given :

• Initial Velocity (u) = 0 m/s

• Acceleration (a) = 0.1 m/s²

• Time taken (t) = 2 minutes = 2 × 60 = 120 s

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To Find :

• (a) The speed acquired = ?

• (b) The distance travelled by bus = ?

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Answer:

⎈ By using the first equation of motion we get :

➦ v = u + at

➦ v = 0 + 0.1 × 120

➦ v = 0 + 12

➦ v = 12 m/s

Therefore, Speed acquired by the bus is 12 m/s.

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⎈ By using third equation of motion we get :

➥ v² - u² = 2as

➥ (12)² - (0)² = 2 × 0.1 × s

➥ 144/2 = 0.1s

➥ 72 = 0.1s

➥ s = 72/0.1

➥ s = 720 m

Therefore, distance travelled by the bus is 720 m.