a car is travelling at a speed of 30km/his brought to a halt in 8m by applying brakes. if the same car is travelling at 60km/h, itcan be bwith the same braking forcerought to a halt
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Case 1:-
Given:-
☆Initial velocity(u)=30km/h
=>30×5/18
=>25/3 m/s
☆Distance(s)=8m
☆Final velocity(v)=0 (as it finally comes to rest)
To find:-
☆Here,we shall find the acceleration to use it in Case 2.
Solution:-
By using,the 3rd equation of motion,we get:-
=>v²-u²=2as
=>0-(25/3)²=2×a×8
=>-625/9=16a
=>a= -625/144 m/s²
Case 2:-
Given:-
☆Initial velocity(u)=60km/h
=>60×5/18
=>50/3 m/s
☆Final velocity(v)=0
☆Accelaration(a)= -625/144 m/s²
To find:-
☆Distance(s)
Solution:-
By,using the 3rd equation of motion,we get:-
=>v²-u²=2as
=>0-(50/3)²=2(-625/144)s
=> -2500/9= -1250/144s
=>s= (-2500×144)/(-1250×9)
=>s=32m
Thus,the required distance is 32m.
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