Question

Q15. Velocity of a train changes from 20 m/s to 25 m/s, when it accelerates at a
rate of 2 m/s?m. Find the distance covered by the train.​

Answer 1

S o l u t i o n :-

Given :

• Initial Velocity (u) = 20 m/s
• Final Velocity (v)= 25 m/s
• Acceleration (a) = 2 m/s^2

To Find :

• Distance covered by the train.

We have to firstly find out time by this formula or equation .

• t = v-u/a or v = u+at

⟹ t = v-u/a

⟹t =25-20/2

⟹t = 5/2

⟹t = 2.5 s

Now, we can find out distance covered by the train using this formula or equation :

⟹ s =ut+1/2 at^2

⟹ s =20×2.5+1/2 × 2×2.5^2

⟹ s = 50+2.5^2

⟹ s = 50+6.25

⟹ s = 56.25 m

Hence , distance covered by the train is 56.25 m.

Note :

• The rate of changing of velocity with respect of time is called acceleration.

Answer 2

♣GIVEN♣

• Initial velocity (u) = 20 m/s

• Final velocity (v) = 25 m/s

• Acceleration (a) = 2 m/s²

♣TO FIND♣

• Distance covered (s)

♣SOLUTION♣

• By the first equation of notion let us find out the time taken to cover that distance :-

$\huge\boxed{\boxed{\bf{\mapsto{v = u + at}}}}$

✯ WHERE :

• v = final velocity

• u = initial velocity

• a = acceleration

• t = time

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❐ Putting all values ❐

➱ 25 = 20 + (2 × t)

➱ 25 = 20 + 2t

➱ 25 - 20 = 2t

➱ 5 = 2t

➱ 2t = 5

➱ t = 5/2

➠ t = 2.5 s

★══════════════════════★

• Finally by the third equation of motion let us find the distance covered by the train :-

$\huge { \boxed {\boxed{ \bf{ \mapsto{s\: =\: ut\: +\: ½\: at² }}}}}$

★══════════════════════★

❐ Putting all values ❐

➱ s = 20 × 2.5 + ½ × 2 × 2.5²

➱ s = 50 + 2.5²

➱ s = 50 + 6.25

➠ s = 56.25 m

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♣ANSWER♣

• The distance covered by the train is 56.25 m.