A car moving with a speed of 40 m/s comes to rest
after travelling a distance 400 m on applying the
brakes. The acceleration of the car is
Answers
Answer:In the above question , the following information is given -
A car moving with a speed of 40 m/s comes to rest after travelling a distance 400 m on applying the brakes .
Solution -
Here , the innitial Velocity of the car is 40 m / s.
So , u = 40 m / s.
The car stops finally .
Hence , the final velocity of the car is 0 m / s .
Now , according to the first equation of motion -
Acceleration = ∆ V / t
Acceleration = ( v - u ) / t
=> Acceleration = ( -40 / t )
According to the second equation of motion ,
S = ut + ( 1 / 2 ) a t ^ 2
Now ,
S refers to the displacement and is equal to 400 metres .
Substituting the given Values -
40t + ( 1 / 2 ) × ( - 40 / t ) A× t^2 = 400
=> 40t - 20t = 400
=> 20t = 400
=> t = 20 seconds. .
Substituting this value -
Acceleration = ∆ V / t
=> ( -40 / 20 ) m s^ ( -2 )
=> -2 m s^ ( -2 )
The negative sign shows retardation as the car is finally stopping ...
Explanation:
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Answer:
6m/s^2
Explanation:
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it's a loda question