the brakes applied to a car produce an acceleration of 6m/s square in the opposite direction to the motion. if the car takes 2 seconds to stopafter the application of brakes, calculate the distanceit travels during this time.
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Ahoy user!
The answer:-
Here, It is clear from the question that,
v= 0m/s (since breaks are applied)
u=?
a = -6m/s^2 (since car moves in the opposite direction)
t = 2seconds
We know,
v= u +at
Now,
=> 0 = u + -6×2
=> 0=u -12
=> u = 12m/s
Now, again we know that
v^2 = u^2 + 2as
=> 0 = (12)^2 + 2× (-6) s
=> 0=144-12s
=> -144=-12s
=> There fore s= -144/-12
= > s = 12 m
THEREFORE,
the distance travelled by the car is 12m in the following time.
Hope this helps!
Cheers ❤
The answer:-
Here, It is clear from the question that,
v= 0m/s (since breaks are applied)
u=?
a = -6m/s^2 (since car moves in the opposite direction)
t = 2seconds
We know,
v= u +at
Now,
=> 0 = u + -6×2
=> 0=u -12
=> u = 12m/s
Now, again we know that
v^2 = u^2 + 2as
=> 0 = (12)^2 + 2× (-6) s
=> 0=144-12s
=> -144=-12s
=> There fore s= -144/-12
= > s = 12 m
THEREFORE,
the distance travelled by the car is 12m in the following time.
Hope this helps!
Cheers ❤
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