Question

A person travelling on a scooter at 43.2 km/h giving a deceleration of 6m/s² to his scooter. How far will it travel before stopping (v=0????​

Answer 1

Answer:

s=12

Explanation:

Here, initial velocity = u = 43.2 km / hour = {(43.2 * 1000) / 3600} m / sec = 12 m / sec.

Final velocity = v = 0 m / sec.

Acceleration = a = -6 m / (sec^2).

Let, S be the distance travelled.

Now, (v^2) = (u^2) + 2*a*S

(0^2) = (12^2) + 2*(-6)*S

12*S = (12^2)

S = 12.

hope it helps you

Answer 2

Given :-

• A person travelling on a scooter at 43.2 km/h giving a deceleration of 6m/s² to his scooter.

To find :-

• How far will it travel before stopping ?

Solution :-

• Initial velocity = 43.2km/h

• Acceleration = - 6m/s²

• Final velocity = 0

→ Final velocity = 43.2km/h

→ v = 12 m/s

• According to third equation of motion

→ v² = u² + 2as

Where " v " is final velocity, " u " is initial velocity, " a " is acceleration and " s " is distance.

• According to the question

→ (0)² = (12)² + 2 × (-6) × s

→ 0 = 144 - 12s

→ 12s = 144

→ s = 144/12

→ s = 12 m

Hence,

• Distance travelled before stopping is 12m

More to know :-

• s = ut + ½ at² (second equation of motion)
• v = u + at (first equation of motion)
• Acceleration is the rate of change in velocity
• Acceleration in negative is know as retardation