Question

A Car Is Moving with A velocity of 10m/s and a constant retardation of 4m/s^2 is acting on it what distance it travel in 3rd second


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Answer 1

Answer:

Explanation:

Given,

Initial velocity, u = 10 m/s

Retardation, a = 4 m/s²

Time taken, t = 3 seconds

To Find,

Distance covered, s = ?

Formula to be used,

1st equation of motion, v = u + at

3rd equation of motion, v² - u² = 2as

Solution,

Putting all the values, we get

v = u + at

⇒ v = 10 + 4 × 3

⇒ v = 10 + 12

⇒ v = 22 m/s

Here, the final velocity is 22 m/s.

Now, the distance covered,

v² - u² = 2as

⇒ (22)² - (10)² = 2 × 4 × s

⇒ 484 - 100 = 8s

⇒ 384 = 8s

⇒ 384/8 = s

⇒ s = 48 m

Hence, the distance covered is 48 m.

Answer 2

Answer:

Given :-

• A car is moving with a velocity of 10 m/s and a constant retardation of 4 m/s² is acting on it.

To Find :-

• What is the distance travelled by a car.

Formula Used :-

❶ To find final velocity we know that,

✪ v = u + at ✪

❷ To find distance covered we know that,

★ v² - u² = 2as ★

where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time
• s = Distance covered

Solution :-

❶ First, we have to find the final velocity,

Given :

• Initial velocity = 10 m/s
• Acceleration = 4 m/s²
• Time = 3 seconds

According to the question by using the formula we get,

⇒ v = 10 + (4)(3)

⇒ v = 10 + 4 × 3

⇒ v = 10 + 12

➦ v = 22 m/s

Hence, the final velocity is 22 m/s .

❷ Now, we have to find the distance covered,

Given :

• Final velocity = 22 m/s
• Initial velocity = 10 m/s
• Acceleration = 4 m/s²

According to the question by using the formula we get,

↦ (22)² - (10)² = 2(4) × s

↦ 484 - 100 = 8 × s

↦ 384 = 8 × s

↦ 384/8 = s

↦ 48 = s

➠ s = 48 m

∴ The distance travelled by a car is 48 m .