Question

A race scooter is seen accelerating uniformly from 18.5m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the scooter and the distance travelled​

Answer 1

Answer :-

→ Acceleration of the sccoter is 11.17 m/s²

→ Distance travelled is 79.81 m .

Explanation :-

We have :-

• Initial velocity (u) = 18.5 m/s

• Final velocity (v) = 46.1 m/s

• Time taken (t) = 2.47 seconds

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We will calculate the acceleration of the race scooter by using the 1st equation of motion.

v = u + at

⇒ 46.1 = 18.5 + a(2.47)

⇒ 46.1 - 18.5 = 2.47a

⇒ 27.6 = 2.47a

⇒ a = 27.6/2.47

⇒ a = 11.17 m/s²

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Now, let's calculate the distance travelled by the scooter using the 3rd equation of motion.

v² - u² = 2as

⇒ (46.1)² - (18.5)² = 2(11.17)s

⇒ 2125.21 - 342.25 = 22.34s

⇒ 1782.96 = 22.34s

⇒ s = 1782.96/22.34

⇒ s = 79.81 m

Answer 2

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★ Question :

• A race scooter is seen accelerating uniformly from 18.5m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the scooter and the distance travelled.

★ Step by step explanation :

• Here we have initial velocity , final velocity of a racer scooter = 18.5m/s , 46.1m/s and time taken = 2.47 seconds.

★ Acceleration of scooter :

➲ㅤㅤㅤv = u + at

➲ㅤㅤㅤ46.1 = 18.5 + a(2.47)

➲ㅤㅤㅤ46.1 - 18.5 = 2.47a

➲ㅤㅤㅤa = 27.6/2.47

➲ㅤㅤㅤa = 11.17 m/s²

•°• Acceleration of scooter = 11.17 m/s²

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━━━━━━

★ Distance travelled by racer scooter :

➲ㅤㅤㅤv² - u² = 2as

➲ㅤㅤㅤ(46.1)² - (18.5)² = 2(11.17)s

➲ㅤㅤㅤ2125.21 - 342.25 = 22.34s

➲ㅤㅤㅤs = 1728.96/22.34

➲ㅤㅤㅤs = 79.81 m

•°• Distance covered by car = 79.81 m

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