Question

a spaceship traveling in space at 300km/s fires it's engine for 15 seconds such that it's final velocity is 600km/s find The
a. acceleration
b. distance travelled in one minute from the time of firing

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Answer 1

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$u = 300 \: kms {}^{ - 1} \\ = 300000 \: ms {}^{ - 1}$

$v = 600 \: kms {}^{ - 1} \\ = 600000 \: ms {}^{ - 1}$

$t = 15 \: sec$

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$a = \frac{v - u}{t}$

$a = \frac{600000 - 300000}{15}$

$a = \frac{300000}{15}$

$\huge\pink{\boxed{a = 20000 \: ms {}^{ - 2}}}$

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$t = 1 \: min \\ = 60 \: sec$

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$distance = speed \times time$

$distance = 600000 \times 60$

$distance = 36000000$

$\huge\pink{\boxed{distance = 3.6 \times 10 {}^{7} m}}$

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Answer 2

Answer:

• Acceleration of Spaceship = 20,000 m/s²
• Distance travelled by Spaceship in one minute from the time of firing = 33,750 km

Explanation:

Given that,

• Initial velocity of spaceship, u = 300 km/s = (300 × 1000) m/s = 300,000 m/s
• Final velocity of spaceship, v = 600 km/h = (600 × 1000) m/s = 600,000 m/s
• Time taken for velocity change, t = 15 sec

We need to find, acceleration of spaceship, a =?

Using First equation of motion

→ v = u + a t

→ 600,000 = 300,000 + a × 15

→ a = ( 600,000 - 300,000 ) / 15

→ a = 300,000 / 15

→ a = 20,000 m/s²

Hence,

Acceleration of spaceship is 20,000 m/s².

Now,

We need to calculate the distance covered by Spaceship in one minute from firing, s =?

Let, distance covered by spaceship in the 15 seconds of accelerating be s₁

so, Using second equation of motion

→ s₁ = 300,000 × 15 + 1/2 × 20,000 × 15²

→ s₁ = 4,500,000 + 2,250,000

→ s₁ = 6,750,000 m = 6,750 km

Now, for next (60-15) seconds means 45 seconds Spaceship will travel with uniform speed of 600,000 m/s, hence acceleration will be zero

Hence, distance covered by Spaceship in these 45 seconds be s₂

→ distance = speed × time

→ s₂ = 600,000 × 45

→ s₂ = 27,000,000 m = 27,000 km

Therefore,

→ total distance covered by spaceship in 1 min = s = s₁ + s₂ = 6,750 km + 27,000 km = 33,750 km.