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Answers
Solution :-
As per the given data ,
- Initial velocity (u) = 10 m/s
- Retardation (a) = - 1.25 m/s ²
- Final velocity (v) = 0 m/s ( as it comes to rest )
As the car is moving with uniform acceleration throughout it's motion we can use the third equation of motion to find distance and first equation of motion to find time
As per the third equation of motion ,
➜ v² = u ² + 2as
On rearranging ,
➜ s = v² - u² / 2a
➜ s = 0 - 100 / - 2 x 1.25
➜ s = - 100 / - 2.5
➜ s = 40 m
The distance covered by the car is 40 m
Now ,
By using first equation of motion ,
➜ v = u + at
On rearranging ,
➜ t = v - u / a
➜ t = 0 - 10 / 1.25
➜ t = 8 sec
The time taken by the car to stop is 8 sec
Given :-
• Initial velocity (u) = 10 m/s
• Retardation (a) = - 1.25 m/s ²
• Final velocity (v) = 0 m/s
To Find :-
• Distance covered, s
• Time taken, t
Solution :-
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1.
2.
3.
Where,
›› s = Distance Covered
›› u = Initial Velocity
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