A bus is moving with a speed of 54 km/have. on applying breaks, it stooped in 8 seconds. calculate the acceleration and distance travelled in this time
Answers
GiveN :-
- initial velocity of bus, u = 54 km h⁻¹
- time taken to stop , t = 8 sec
- final velocity , v = 0
To FinD :-
- Acceleration , a = ?
- distance traveled in this time ,s =?
FormulaE :-
▶ First equation of motion
( where v is the final velocity , u is the initial velocity , a is the acceleration and t is the time taken )
▶ Second equation of motion
( where s is the distance traveled , t is the time , u is the initial velocity , a is the acceleration )
SolutioN :-
Converting initial velocity given in km h⁻¹ into m s⁻¹
initial velocity , u = 54 km h⁻¹
Finding acceleration
Using first equation of motion
putting known values
Finding distance traveled after applying break
Using second equation of motion
putting known values
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Question :-
▪ A bus is moving with a speed of 54 km/have. on applying breaks, it stooped in 8 seconds. calculate the acceleration and distance travelled in this time.
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Given:-
- Initial velocity (u) =54km/h=54×5/18=15m/s
- Final velocity (v) =0m/s
- Time taken =8s
To Find :-
- Acceleration
- Distance travelled
Solution :-
By using 1st equation of motion
➦ v =u+at
➭ 0=15+a ×8
➭ a = -15/8
➭ a = -1.875
here, negative symbol show retardation.
∴The acceleration of bus 1.875m/s².
And,
By using 2nd equation of motion
➦ s =ut +1/2at²
➭ s = 15×8+1/2 ×1.875× 8× 8
➭ s= 7200m