A scooterist moving on a straight highway at 54 km/ h, sights obstacle at a distance at a distance pf 50 m, and applies to produce -2.5 m/sec^2 the retardation. Will he be able to avoid the accident?
Answers
Given:-
- Initial velocity,u = 54km/h
- Final velocity ,v = 0m/s
- Acceleration ,a = -2.5m/s²
To Find:-
- Distance covered ,s
Solution:-
According to the Question
It is given that scooterist moving on a straight highway at 54 km/ h, sights obstacle at a distance at a distance pf 50 m, and applies to produce -2.5 m/sec^2 the retardation.
conversion of unit
→ u = 54×5/18 = 15m/s
firstly we calculate the distance covered by scooterist .
using 3rd equation
- v² = u² + 2as
where,
- v is the final velocity
- a is the acceleration
- u is the initial velocity
- s is the distance covered
Substitute the value we get
→ 0² = 15²+ 2(-2.5) × s
→ 0 = 225 + (-5)s
→ -225 = -5s
→ 225 = 5s
→ s = 225/5
→ s = 45
Hence the distance covered by the scooterist is 45 metres.
Since, the obstacle was at 50m from the scooterist but the distance covered by scooterist was 45 metres . Then he will be able to avoid the accident .
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Given :-
A scooterist moving on a straight highway at 54 km/ h, sights obstacle at a distance at a distance pf 50 m, and applies to produce -2.5 m/sec^2
To Find :-
Will he be able to avoid the accident?
Solution :-
We know that
1 km/h = 5/18 m/s
54 km/h = 54 × 5/18 = 3 × 5 = 15 m/s
Now
v² - u² = 2as
0² - 15² = 2(-2.5)(s)
0 - 225 = -5s
-225 = -5s
-225/-5 = s
225/5 = s
45 = s
Given distance is 50 m.
45 m is less than 50 m. Therefore, he may avoid acciednt