Question

A racing car has a uniform acceleration of 4 metre per second square.What distance will it cover in 10 s after start?​

Answer 1

Answer:

• The distance travelled (S) is 200 m

Given:

1. Acceleration (a) = 4 m/s²
2. Time given (t) = 10 seconds.
3. Initial velocity (u) = 0 m/s

Explanation:

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From, second kinematic equation we know,

⇒ S = u t + 1/2 a t²

Here,

• S Denotes Distance travelled.
• u Denotes initial velocity.
• t Denotes time taken.
• a Denotes acceleration.

Substituting the values,

⇒ S = 0 × 10 + 1/2 × 4 × (10)²

⇒ S = 0 + 1/2 × 4 × 100

⇒ S = 1/2 × 4 × 100

⇒ S = 2 × 100

⇒ S = 200

⇒ S = 200 m

∴ The distance travelled (S) is 200 m.

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Alternate Method:

From first kinematic equation, we know,

⇒ v = u + a t

Here,

• v Denotes Final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• t Denotes Time taken.

Substituting the values,

⇒ v = 0 + 4 × 10

⇒ v = 4 × 10

⇒ v = 40

⇒ v = 40 m/s

∴ We got the final velocity.

Now, Applying third kinematic equation

⇒ v² = u² + 2 a S

Here,

• v Denotes Final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• S Denotes Distance travelled.

Substituting the values,

⇒ (40)² = (0)² + 2 × 4 × S

⇒ 1600 = 0 + 8 × S

⇒ 1600 = 8 S

⇒ S = 1600/8

⇒ S = 200

⇒ S = 200 m

∴ The distance travelled (S) is 200 m.

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Answer 2

Hola ⚘⚘

Answer:

200m

Explanation:

Given: Acceleration = 4m/s²;Time= 10s; Initial velocity = 0m/s

>> s = ut + 1/2 at²

• s = distance travelled

• u = Initial velocity

• t = time taken

• a = acceleration

>> s = 0*10+1/2*4*10²

>> s = 1/2*4*100

>> s = 0.5*4*100

>> s = 0.5*400

>> s = 200m