Question

A particle covers 10m in first 5 sec and 10m in next 3sec.Assuming constant acceleration ,find initial speed,acceleration and distance covered in next 2 secs.(7/6 m/s,1/3 m/s ,8.33 m)..
guys plz answer ​

Answer 1

Given :-

▪ A particle covers 10 m in first 5 seconds , and 10 m in next 3 seconds. Acceleration is constant.

To Find :-

▪ Initial speed, Acceleration and distance covered in next 2 seconds.

Solution :-

We are given, time taken = 5 seconds and displacement made = 10 m.

Using 2nd equation of motion, acceleration can be calculated as,

⇒ s = ut + 1/2 at²

⇒ 10 = u×5 + 1/2 × a × 5²

⇒ 10 = 5u + 25a / 2

⇒ 20 = 10u + 25a

⇒ 4 = 2u + 5a ...(1)

Now, In the next 3 seconds It would cover 10 m,

So, we have

• Displacement made = 10 + 10 = 20 m
• time taken = 5 + 3 = 8 seconds

Using 2nd equation of motion, we have

⇒ s = ut + 1/2 at²

⇒ 20 = u×8 + 1/2×a×8²

⇒ 20 = 8u + 32a

⇒ 5 = 2u + 8a ...(2)

Subtract (1) from (2), we get

⇒ 2u + 8a - 2u - 5a = 5 - 4

⇒ 3a = 1

⇒ a = 1/3 m/s²

Now, Substitute a = 1/3 in (1), we get u

⇒ 2u + 5×1/3 = 4

⇒ 2u = 4 - 5/3

⇒ 2u = (12 - 5)/3

⇒ 2u = 7/3

⇒ u = 7/6 m/s

Now, Let us find the distance covered in next 2 seconds.

The distance covered in the next seconds is the difference of distance covered in 8 seconds and the distance covered in (8+2) = 10 seconds.

So, We have

⇒ Distance travelled = (ut + 1/2at²) - (ut₂ + 1/2at₂²)

⇒ d = 20 - (7/6 × 10 + 1/2 × 1/3 × 100)

⇒ d = 20 - (70/6 + 1/600)

⇒ d = 20 - (7000 + 1)/600

⇒ d = 20 - 7001/600

⇒ d = (12000 - 7001)/600

⇒ d = 4999 / 600

⇒ d = 8.33 m