10. The breaks applied to a car produce a
negative acceleration of 60 m/s. If the
car takes 2s to stop after applying
the breaks, Calculate the distance it
travels during this time.
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10. The breaks applied to a car produce a
negative acceleration of 60 m/s. If the
car takes 2s to stop after applying
the breaks, Calculate the distance it
travels during this time.
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Answer:
Given :-
acceleration = ( -60m/s^2 )
time = 2s
initial velocity = 0m/s [ as taken ]
To find :-
distance = ?
Formula :-
S = ut + 1/2at^2 _______equation ( 2 ) [ Motion ]
Solution :-
S = 0m/s × 2s + 1/2 × ( -60m/s^2 ) × ( 2s )^2
= 1/2 × ( -60m/s^2 ) × 4s^2
= 1/2 × ( -60m ) × 4
= -120m