Question

an express train moving to 30m/s reduces its speed to 10m/s in a distance of 200m if the breaking force is increased by 12.5% in the beginning find the distance that it travels before coming to rest.

Answer 1

Given :

The initial speed of train = u = 30 m/s

The reduced speed of train = v = 10 m/s

The distance cover by train =s  = 200 m

The increased breaking force applied to reduced speed = 12.5 %

To Find :

The distance that it travels before coming to rest.

Solution :

From Newton's motion equation

 v² = u² + 2 a s

where   v = finial speed

             u = initial speed

            a = acceleration

             s = distance cover

So,    (10 m/s)² = (30 m/s)² + 2 × a ×  200 m

Or,     100 = 900 + 400 a

Or,   400 a = 100 - 900

Or,    400 a = - 800

∴               a = $\dfrac{-800}{400}$

i.e acceleration = - 2 m/s²

or, retardation =  2 m/s²

Now, Breaking force increase by 12.5 %

So, New acceleration = $\left | a_n_e_w \right |$ = 2 +  2 × 12.5 %

                                               =  2 + 0.25

                                                = 2.25 m/s²

So, new acceleration = - 2.25 m/s²

This acceleration is applied in opposite direction to stop train

Now, Again

From Newton's motion equation

      v² = u² + 2 a s'

where   v = finial speed = 0 m/s²

             u = initial speed = 30 m/s

            a = acceleration = - 2.25 m/s²

             s' = distance cover

So,    (0 m/s)² = (30 m/s)² + 2 × (  - 2.25 m/s²)  × s'

Or,     0 = 900 - 4.5 s'

or,     4.5 s' = 900

∴             s' = $\dfrac{900}{4.5}$

i.e           s' = 200 meters

So, Train will stop by travelling distance of s' = 200 meters

Hence, The distance that train travels before coming to rest is 200 meters  Answer

Answer 2

Given:

(i) Express train is moving at 30m/s

(ii) Speed is reduced to 10m/s in a distance of 200m, when the breaking force is increased by 12.5% in the begining.

To find:

(i) The distance travelled before coming to rest.

Solution:

Given, the initail velocity (u) of the train = 30m/s

The final velocity (v) of the train = 10m/s

Distance covered while braking, that is displacement(S) = 200m

According to Newton's laws of motion,

v² = u² + 2aS

⇒ 10² = 30² + 2a(200)

⇒ 400a = 100-900

⇒ a = -800/400

⇒ a = -2 m/s²

Again, when the braking force is increased by 12.5%, we have to find distance travelled before coming to rest.

Magnitude of acceleration

= (100+12.5)% of 2m/s²           [Negative sign is ignored as we are calculating magnitude only]

= 112.5% of 2

= 2.25 m/s²

Now, the new stopping distance (S') is given as:

v² = u² + 2aS'

Now, v = 0 as the train will come to haltand u = 30m/s

⇒ 2aS' = 0² - 30²

⇒ 2*(-2.25)*S' = -900

⇒ -4.5S' = -900

⇒  S' = -900/-4.5

⇒  S' = 200m

The train will travel 200m before coming to rest.