a car moving with speed of 50 km/hr,can be stopped by brakes after at least 6m. if the same car is moving at a speed of 100km/hr,the minimum stopping distance is
Answers
Answered by
326
Given
Speed = 50 km/hr
Speed of car= 50 km/hr
when brakes applied final velocity become zero,
So we apply third equation of motion
v²- u² = 2 as
where u = Initial value = 50 km/hr
v = Final value = 0
a= acceleration
s = Distance = 6 m
Applying the value in formulae
v²- 50² = 2 a × 6
a = -2500 /12
To find minimum stopping distance
v² = 2 a s + u²
100² = 2 × (-2500/12 ) s
100 ×100 × 6 / 2500 = s
24 m = s
Answered by
108
case 1 :- speed of car, u = 50km/h = 50 × 5/18 = 125/9 m/s
a/c to question, car can be stopped by brakes after at least 6m.
so, stopping distance of car, S = 6m
we know formula of stopping distance = u²/2a, where u is initial speed of car and a is retardation acts on car.
now, 6 = (125/9)²/2a
a = (125/9)²/12 ...........(1)
case 2 :- speed of car , U = 100km/h = 100 × 5/18
= 250/9 m/s
so, stopping distance, S' = (250/9)²/2a
= (250/9)²/2{(125/9)²/12}
= 6 × 4 = 24m
hence, minimum stopping distance of 2nd car is 24m
a/c to question, car can be stopped by brakes after at least 6m.
so, stopping distance of car, S = 6m
we know formula of stopping distance = u²/2a, where u is initial speed of car and a is retardation acts on car.
now, 6 = (125/9)²/2a
a = (125/9)²/12 ...........(1)
case 2 :- speed of car , U = 100km/h = 100 × 5/18
= 250/9 m/s
so, stopping distance, S' = (250/9)²/2a
= (250/9)²/2{(125/9)²/12}
= 6 × 4 = 24m
hence, minimum stopping distance of 2nd car is 24m
Similar questions
Final velocity, v = 0
Distance travelled before coming to rest, s = 6 m
Using, v2 = u2 + 2as
=> a = -u2/(2s)
=> a = -16.075 m/s2
Again,
u = 100 km/h = 500/18 m/s
v = 0
a = -16.075 m/s2
Now, v2 = u2 + 2as
=> s = -u2/(2a) = 24 m