Question

A train is travelling at a speed of

90 km h–1. Brakes are applied so

as to produce a uniform

acceleration of – 0.5 m s-2. Find

how far the train will go before it

is brought to rest.

​

Answer 1

Answer:

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s

acceleration , a = -0.5 m/s²

Use formula,

v = u + at

Finally train will be rest so, final velocity, v = 0

0 = 25 - 0.5t

25 = 0.5t ⇒t = 50 sec

Again, use formula,

S = ut + 1/2at²

Where S is distance travelled before stop

S = 25 × 50 - 1/2 × 0.5 × 50²

= 1250 - 1/2 × 0.5 × 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec

Answer 2

GIVEN:

• Initial Velocity (u) of train = 90 km/hr
• Final velocity (v) = 0 m/s (Breaks applied)
• Acceleration (a) = -0.5 m/s²

TO FIND:

• What is the distance covered by the train ?

SOLUTION:

We have given that, the initial Velocity of the train is 90 km/hr

Convert km/hr into m/s:-

➫ $\rm{ u = 90 km/hr}$

➫ $\rm{ u = \cancel{90} \times \dfrac{5}{\cancel{18}}}$

➫ $\rm{ u = 5 \times 5}$

➫ $\bf{ u = 25 \: m/s}$

Let the time taken by the train be 't' seconds

To find the time taken, we use the First equation of motion:-

$\large\bf{\star \: v = u + at \: \star}$

According to question:-

$\rm{\dashrightarrow 0 = 25 + (-0.5) \times t }$

$\rm{\dashrightarrow 0 = 25 - 0.5t }$

$\rm{\dashrightarrow 0.5t = 25 }$

$\rm{\dashrightarrow t = \cancel\dfrac{25}{0.5} }$

$\bf{\dashrightarrow t = 50 \: seconds }$

❍ Now, we have to find the distance covered by the train

To find the distance covered by the train, we use the second equation of motion:-

$\large\bf{\star \: s = ut + \dfrac{1}{2} at^2\: \star}$

According to question:-

$\rm{\dashrightarrow s = 25 \times 50 + \dfrac{1}{2} \times (-0.5) \times (50)^2 }$

$\rm{\dashrightarrow s = 1250 + \dfrac{1}{\cancel{2}} \times (-0.5) \times \cancel{2500}}$

$\rm{\dashrightarrow s = 1250 - 0.5 \times 1250 }$

$\rm{\dashrightarrow s = 1250 - 625 }$

$\bf{\dashrightarrow \star \: s = 625 \: m \: \star}$

❝ Hence, the distance covered by the train is 625 m ❞

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