Question

a bus moving with speed of 60 km per hour can be stopped by applying brakes after at least 3 M if the same bus moving with a speed of 80 km per hour what is the minimum stopping distance

Answer 1

Here is your answer.

First convert 60 km/hr to m/s

==> 60*5/18 = 50/3 m/s

the final velocity of the bus would be 0 after applying the brakes.

Displacement = 3 m

Applying

v² - u² = 2as

0² - (50/3)² = 2*a*3

- 2500/9 = 6*a

- 277.7 = 6*a

a = -277.7 ÷ 6

a = - 46.2 m/s² (retardation)

Now,
speed = 80 km/hr

In m/s

=>80*5/18 = 200/9

Finding minimum stopping distance

s = v² - u² / 2a

s = (200/9)² - 0² / 2*46.2

s = 40000/81 / 92.4

s = 493.8 / 92.4

s = 5.344 m

I HOPE THE ANSWER IS CORRECT AND HELPS YOU

Answer 2

The minimum stopping distance is 5.33 m.

A bus moving with speed of 60 km per hour can be stopped by applying brakes after at least 3m if the same bus moving with a speed of 80 km per hour.

We have to find the minimum stopping distance.

We know, stopping distance can be found by,

$S=ut+\frac{u^2}{2a}$

minimum stopping distance is possible only when , t = time of reaction = 0

$\implies S=\frac{u^2}{2a}$

Where, u = initial speed

a = retardation applied on the bus to stop it.

S = total distance covered until it stops.

Given,

S₁ = 3m, u₁ = 60km/h , u₂ = 80 km/hr

⇒ S₁/S₂ = u₁²/u₂² [ a remains same for both cases ]

⇒ 3/S₂ = (60/80)² = 9/16

⇒ S₂ = 16/3 = 5.33 m

Therefore the minimum stopping distance is 5.33 m.

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