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
- 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
Since the obstacles are at the distance of 50 m
So he/she able to avoid himself
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
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