Question

An object starts from rest with a constant acceleration of 5 ms-2 for 5 5 seconds and then move with constant velocity. Find the distance travelled by it in last two seconds of its accelerated motion.
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Answer 1

Answer:

Initially the object moves with constant acceleration of 5 m/s² for 5 seconds. Therefore the distance covered by the object in this 5 second interval is:

⇒ s = ut + 0.5 at²

⇒ s = (0 × 5) + 0.5 × 5 × 5²

⇒ s = 62.5 m

Therefore the object has travelled a total of 62.5 m in 5 seconds.

We are required to find the distance travelled by the object in last 2 seconds of it's constant acceleration phase. This can be calculated as:

⇒ Distance in last 2 seconds = Distance in 5 seconds - Distance in 3 seconds

Distance covered in first 3 seconds is:

⇒ s = ut + 0.5 at²

t = 3 seconds, hence we get:

⇒ s = (0 × 3) + 0.5 × 5 × 3²

⇒ s = 22.5 m

Hence Distance covered in last 2 seconds of it's travel is:

⇒ Distance covered = 62.5 m - 22.5 m

⇒ Distance covered = 40 m

Hence Option (B) is the correct answer.

Answer 2

Answer:

Question :-

✯ An object starts from rest with a constant acceleration of 5 ms-2 for 5 5 seconds and then move with constant velocity. Find the distance travelled by it in last two seconds of its accelerated motion.

Given :-

✯ An object starts from rest with a constant acceleration of 5 ms-2 for 5 5 seconds and then move with constant velocity.

Find Out :-

✯ Find the distance travelled by it in last two seconds of its accelerated motion.

Solution :-

✭ In first case :-

As we know that :

☆ $\red{ \boxed{\sf{s =\: ut + \dfrac{1}{2}at^2}}}$ ☆

➙ $\bf s =\: ut + \dfrac{1}{2}at^2$

➙ $\sf s = 0 \times 5 + \dfrac{1}{2} \times 5 \times (5)^2$

➙ $\sf s = 0 \times 5 + \dfrac{1}{2} \times 5 \times (5 \times 5)$

➙ $\sf s = 0 \times 5 + \dfrac{1}{2} \times 5 \times 25$

➙ $\sf s = 0 \times 5 + \dfrac{1}{2} \times 125$

➙ $\sf s = 0 + 62.5$

➙ $\sf\boxed{\bold{\red{s = 62.5\: m}}}$

Henceforth, the object travelled 62.5 m.

✭ In second case :-

So, distance covered in first 3 seconds is :-

➙ $\bf s =\: ut + \dfrac{1}{2}at^2$

➙ $\sf s = 0 \times 3 + \dfrac{1}{2} \times 5 \times (3)^2$

➙ $\sf s = 0 \times 3 + \dfrac{1}{2} \times 5 \times (3 \times 3)$

➙ $\sf s = 0 \times 3 + \dfrac{1}{2} \times 5 \times 9$

➙ $\sf s = 0 \times 3 + \dfrac{1}{2} \times 45$

➙ $\sf s = 0 + 22.5$

➙ $\sf\boxed{\bold{\green{s = 22.5\: m}}}$

Hence, the distance travelled by it in last two seconds of its accelerated motion :-

➙ Distance Travelled = 62.5 m - 22.5 m

➙ ${\small{\bold{\purple{\underline{Distance\: Travelled = 40\: m}}}}}$

Henceforth, the distance travelled by it in last two seconds of its accelerated motion is 40 m .