Question

A bus is moving with a speed 72 km/h can be stopped by brakes
after atleast 10 m. What will be the minimum stopping distance, if the
same bus is moving at a speed of 144 km/h?





please answer
​

Answer 1

Answer:

☆Hey Mate☆

Question⤵

A bus is moving with a speed 72 km/h can be stopped by brakes after atleast 10 m. What will be the minimum stopping distance, if the same bus is moving at a speed of 144 km/h?

Answer⤵

Distance 2aS = v^2-u^2

Where v=final velocity and u=initial velocity;

taking S=10m, and u = 72km/h;

20a=0-5184

Therefore,

a= -5184/20

= -259.2 m/s^2

So when u = 144 km/h,

2*259.2*S =20736

and S = 20736/518.4

=40m

Hope It Helpss...!!

Answer 2

Given :-

Distance covered = 10 m

Initial velocity of the bus = 72 km/h

Final velocity of the bus = 0 m/s

To Find :-

The minimum stopping distance, if the  same bus is moving at a speed of 144 km/h.

Analysis :-

Here we are given with the distance, initial and final velocity.

Firstly find the acceleration by substituting the values from the question using the third equation of motion.

Then by using the third equation of motion find the minimum stopping distance accordingly.

Solution :-

We know that,

• u = initial velocity
• a = Acceleration
• v = Final velocity
• s = Displacement

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2-u^2=2as }}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 72 km/h = 20 m/s

Displacement (s) = 10 m

Substituting their values,

⇒ 0² - 20² = 2 × a × 10

⇒ 0 - 400 = 20a

⇒ -400 = 20a

⇒ a = -400/20

⇒ a = -20 m/s

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2-u^2=2as }}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 144 km/h = 40 m/s

Acceleration (a) = -20 m/s

Substituting their values,

⇒ 0² - 40² = 2 × -20 × s

⇒ 0 - 1600 = -40 × s

⇒ -1600 = -40 × s

⇒ s = -1600/-40

⇒ s = 40 m

Therefore, the minimum stopping distance is 40 m.