Question

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.​

Answer 1

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