A car moving with a velocity of 30 m/s is sttoped by the application of brakes which impact a retardation of 6m/s² to the car.
(i) How long does it take for the car to come to a stop.
(ii) How far does the car trqvel during the time breaks are applies ?
Answers
Answered by
3
Answer:
(i)5 seconds (ii)75 meters
Explanation:
u=30m/s
v=0
a= -6m/s²
(i)a=v-u/t
t=v-u/a
=0-30/-6
5sec
(ii)s=ut+0.5at²
=30×5 - ½×6×5×5
=150-75
=75m
PLEASE MARK BRAINIEST
Answered by
4
Explanation:
Given -
u = 30 m/s (initial velocity)
v = 0 m/s (final velocity)
a = -6 m/s² (retardation)
To Find -
- How long does it take for the car to come to a stop.
- How far does the car travel during the time breaks are applies ?
As we know that :-
- v = u + at
» 0 = 30 + (-6)t
» -30 = -6t
» t = -30/-6
- » t = 5 seconds
And
As we know that :-
- s = ut + 1/2at²
» s = 30×5 + 1/2×-6×(5)²
» 150 - 75
- » 75 m
Hence,
Car take 5 seconds to stops
and
Car travels 75 m during the time breaks are applies.
Formula Used :-
- s = ut + ½at²
- v = u + at
here,
v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance travelled
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