Question

A train starts from rest. It moves through 1 Km in 100 s with uniform acceleration.
What will be its speed at the end of 100 s.​

Answer 1

Given :

Initial velocity, u = 0 m/s (as it is starts from rest)

Distance, s = 1 km = 1000 m

Time, t = 100 seconds

To find :

Speed, v

According to the question,

➞ s = ut + ½ at²

Where,

s = Distance

a = Acceleration

t = Time

u = Initial velocity

➞ 1000 = 0 × ½ 100 + a × 100 × 100

➞ 1000 = 0 + 5000a

➞ 1000 - 0 = 5000a

➞ 1000 = 5000a

➞ 1000 ÷ 5000 = a

➞ 0.2 = a

So,the acceleration is 0.2 m/s².

Now,

➞ v = u + at

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

➞ v = 0 + 0.2 × 100

➞ v = 0 + 20

➞ v = 20

So,the speed is 20 m/s.

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Answer 2

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First using 2nd equation of motion -

$\large \: S \: = \: ut+ \frac{1}{2} \: at^2$

$\large \: ➟ 1000=0×100 + \frac{1}{2} a (100)^2$

$\large \: ➟ \: 1000=5000a$

$\large \: a= \frac{1}{5} \: m/s^2$

Now we need to find the final velocity ‘v’ as asked in the question

So by using 1st equation of motion, We get :-

$\large \: V=u + at$

$\large \: ➟ \: V= 0 + \frac{1}{5} × 100$

$\large \: ➟ \: V=20m/s$

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