Physics, asked by krupa212010106, 6 hours ago

A scooterist travelling at a speed of 54km/h sights an obstacle 50m away. He applies brakes to avoid the accident. Find the retardation produced.

Class 9th Physics chapter 8 Motion

Answers

Answered by SHUBHMALHOTRA
8

 \large \bf \underline\purple{Given \:  From  \: Question ❓}

  • Initial Velocity , [u] = 54 Km/h
  • Final Velocity , [v] = 0 m/s

We have to find distance .

Let's take distance to be X

According to the question .

The scooterist moving on staight highway at

54 Km/h

Obstacle at a distance is = 50 M

Conservation of unit .

U = 54 × 5/18 = 15 m.

By using 3rd equation .

= + 2as

  • V is final Velocity
  • a is acceleration
  • U is the initial velocity
  • S distance cover

O² = S²+2(-2.5) × 5

☞ 0 = 225 + (-5) S

-225 = (-5)

225 = 5

S = 225/5

 \large \bf \underline\pink{45 \: meters }

Since the obstacles are at the distance of 50 m

So he/she able to avoid himself

Answered by sumellikaagnisha
1

Given From Question❓

Initial Velocity , [u] = 54 Km/h

Final Velocity , [v] = 0 m/s

We have to find distance .

Let's take distance to be X

According to the question .

The scooterist moving on staight highway at

54 Km/h

Obstacle at a distance is = 50 M

Conservation of unit .

U = 54 × 5/18 = 15 m.

By using 3rd equation .

V² = u² + 2as

V is final Velocity

a is acceleration

U is the initial velocity

S distance cover

O² = S²+2(-2.5) × 5

☞ 0 = 225 + (-5) S

-225 = (-5)

225 = 5

S = 225/5

\large \bf \underline\pink{45 \: meters }

Since the obstacles are at the distance of 50 m

So he/she able to avoid himself

Initial Velocity , [u] = 54 Km/h

Final Velocity , [v] = 0 m/s

We have to find distance .

Let's take distance to be X

According to the question .

The scooterist moving on staight highway at

54 Km/h

Obstacle at a distance is = 50 M

Conservation of unit .

U = 54 × 5/18 = 15 m.

By using 3rd equation .

V² = u² + 2as

V is final Velocity

a is acceleration

U is the initial velocity

S distance cover

O² = S²+2(-2.5) × 5

☞ 0 = 225 + (-5) S

-225 = (-5)

225 = 5

S = 225/5

[tex]

\large {\bf {\underline{\pink{45 \: meters }}}} [\tex]

Since the obstacles are at the distance of 50 m

So he/she able to avoid himself

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