Physics, asked by aksingodiya, 1 year ago

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


akashbond777p7ixil: Initial velocity, u = 50 km/h = 50 × (5/18) = 250/18 m/s

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

Answers

Answered by lodhiyal16
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 abhi178
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
Similar questions