Question

an Express train moving at 30 M per second reduces its speed at 10 M per second in a distance of 240 M if the braking force is increased by 12.5% in the beginning find the distance that it travels before coming to rest

Answer 1

Heya

The equation of motion to use is :

V² = U² + 2as

V = 10m/s U = 30m / s S= 240m

Doing the substitution we have :

100 = 900 + 480a

- 800 = 480a

a = - 5/3 m/s²

Basic definition of force

F = ma

Thus a = F/m = -5/3 m/s²

a is directly proportional to F.

An increase in F by x increases the acceleration by x also.

Since our breaking force is increased 12.5% then the new acceleration is:

a (new) = (1 + 0.125) × -5/3 m/s² = -15/8 m/s²

Using the same original equation of motion and solve for S

S = (V² - U²) / (2a)

Substituting:

this time V = 0 m/s (the train comes to rest)

S = ((0) - (900) / (2 × -15/8 m/s²) = - 900 ÷ - 15/8

900 × 8 / 15 = 240 m

Answer 2

hey  

gd morng

here is your answer

given that the train is moving initially at 30m/s and then decelerates to 10m/s then

v2_u2=2as

(10)2-(30)2=2(-a)(240)

100-900=(-480a)

-800= -480a

a= -5/3m/s2

again tis given that if initially the breaking force is 12.5%

the by f=ma, as f increases ,aceleration also increases,therefore

new aceleration is a( 1+12.5/100 )(-50/3)

a=1.875 or 15/8m/s2

now use the sformula u2/2a

            d = (30)2/2*(-15/8)m

              d = 240m

therefore the train will come to rest or stom at the ditance of 240m if the breaking force is 12.5% in the initial.

hope its help you

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