A car a moving at rate of 72km/h and applies brakes which provide a retardation of 5ms-2 . (i) How much time does the car takes to stop. (ii) How much distance does the car cover before coming to rest? (iii) What would be the stopping distance needed if speed of the car is doubled?
Answers
Answered by
156
u = 72 km/h
= 72×5/18
= 20 m/s
a = -5 m/s²
(i) v = 0
v = u+at
0 = 20-5t
-5t=-20
t = 20/5
t = 4 s
(ii) s = ut+1/2at²
= 20×4 + 1/2×-5×4²
= 80 - 1/2×5×16
= 80 - 40
= 40 m
(iii) Speed is doubled :-
u = 20×2 = 40 m/s
v = 0 m/s
a = -5 m/s²
v²-u²=2as
0²-40²=2×-5s
-1600=-10s
s = -1600/-10
s = 160 m
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= 72×5/18
= 20 m/s
a = -5 m/s²
(i) v = 0
v = u+at
0 = 20-5t
-5t=-20
t = 20/5
t = 4 s
(ii) s = ut+1/2at²
= 20×4 + 1/2×-5×4²
= 80 - 1/2×5×16
= 80 - 40
= 40 m
(iii) Speed is doubled :-
u = 20×2 = 40 m/s
v = 0 m/s
a = -5 m/s²
v²-u²=2as
0²-40²=2×-5s
-1600=-10s
s = -1600/-10
s = 160 m
Hope it helps you ☺
Please mark as Brainliest ☺
Answered by
27
The time taken is
As it is given in kmph
So to convert it into SI UNIT
72*5/18 = 20 m/sec
Time taken is 20/5 = 4 seconds
20*20 = 10s
400/10 = 40 metres
Now if the speed is doubled
Then velocity would be 40 instead of 20
Then
40*40/10 = s
1600/10 = s
160 meters
As it is given in kmph
So to convert it into SI UNIT
72*5/18 = 20 m/sec
Time taken is 20/5 = 4 seconds
20*20 = 10s
400/10 = 40 metres
Now if the speed is doubled
Then velocity would be 40 instead of 20
Then
40*40/10 = s
1600/10 = s
160 meters
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