Question

A train increases its speed from 18 km/h to 81 km/h in half a minute. What was its acceleration? What distance did it cover in this half-minute?

Answer 1

Given

Initial velocity of train = 18 km/h

Final velocity of train = 81 km/h

Time taken = half minute

To Find

i) Acceleration of train

ii) Distance covered in half minute

Solution

First we will convert velocities from km/h to m/s

Here,initial velocity = 18 km/h

u = 18 × (5/18) m/s

u = 5 m/s

Again, final velocity = 81 km/h

v = 81 × (5/18) m/s

v = 405/18

v = 22.5 m/s

Here, time = half minute = ½(60) = 30 seconds.

Now using equation of Kinematics :

⇒ a = (v - u)/t

⇒ a = (22.5 - 5)/30

⇒ a = 17.5/30

⇒ a = 0.58 m/s²

So, acceleration of train = 0.58 m/s² (i)

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Now we know equation of Kinematics :

⇒ s = ut + ½at²

⇒ s = 5(30) + ½(0.58 × 30²)

⇒ s = 150 + ½(0.58 × 900)

⇒ s = 150 + ½(522)

⇒ s = 150 + 261

⇒ s = 411 m

So, distance covered by train = 411 m (ii)

Answer 2

GIVEN :-

A train increases its speed from 18 km/h to 81 km/h in half a minute.

TO FIND :-

✺ The acceleration of the train.

✺ The distance covered by the train in half a minute.

SOLUTION :-

✺ Initial velocity(u) = 18 km/h = 18*(5/18) = 5 m/s

✺ Final velocity(v) = 81 km/h = 81*(5/18) = 22.5 m/s

✺ Time(t) = 1/2 minute = 30 secs

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As we know that,

$\displaystyle{\bold{v=u+at}}$

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Now, putting the values :-

22.5 = 5 + a(30)

⇨ 17.5 = 30a

⇨ a = 17.5/30

⇨ a = 0.58 m/s²

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We also know that,

$\displaystyle{\bold{v^2-u^2=2as}}$

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Now, putting the values :-

(22.5)² - (5)² = 2 × 0.58 × s

⇨ 506.25 - 25 = 1.16s

⇨ 481.25 = 1.16s

⇨ s = 481.25/1.16

⇨ s = 414.87 m

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