Question

A train is travelling at speed of 75 kilometre per hour the brakes are applied so as to produce a uniform acceleration of 2.5 metre per second square find how far the train goes before it stops

Answer 1

Answer :

• The distance covered by the train before coming to rest , S = 15 m.

Explanation :

Given :

• Intial velocity of the train = 75 km/hr.

• Final velocity of the train = 0

[Here the final velocity is taken as zero because the breaks are applied]

• Acceleration of the train = -2.5 m/s²

[Acceleration produced will be negative since the train is retarding (uniformly)]

To find :

• Distance covered by train before coming to rest.

Knowledge required :

let us convert the initial velocity from km/h to m/s.

To convert from km/hr , we have to find the product of the speed and 5/18.

So by using this information , we get :

==> 75 km/h

==> (75 × 5/18) m/s

==> 375/18 m/s

==> 20.84 (approx.)

==> 20 m/s

∴ 75 km/h = 20 m/s.

Hence the initial velocity of the train is 20 m/s.

Solution :

We know the third equation of motion i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² ± 2aS

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• S = Distance

Now by using the third equation of motion and substituting the values in it, we get :

==> v² = u² ± 2aS

==> v² = u² - 2aS

==> 0² = u² - 2aS

==> 0 = u² - 2aS

==> -u² = - 2aS

==> u² = 2aS

==> 20² = 2 × 2.5 × S

==> 400 = 5S

==> 400/5 = S

==> 80 = S

∴ S = 80 m.

Therefore ,

• The distance covered by train before coming to rest is 80 m.

Answer 2

$\bold{Quêstioñ }$

A train is travelling at speed of 75 kilometre per hour the brakes are applied so as to produce a uniform acceleration of 2.5 metre per second square find how far the train goes before it stops

$\bold{Solution : }$

Vinod Vinod the third equation motion i.e,

v²=u²±2aS

$\bold{where : }$

V=Final velocity.

U=Inihal velocity.

A=Acceleration.

S=distance.

$\bold{ANSWER : }$

=>V²=U²±2aS

=>V²=U²-2aS

=>O²=U²-2aS

=>O=U²-2aS

=>-U²=-2aS

=>U²=2aS

=>20²=2×2.5×S

=>400=5S

$\bold{=80=S}$

$\bold{there fore=S=80m}$