Question

A body moving with uniform acceleration describes 10 metre in the 3rd sec and 12 metre in the 5th sec.

Find the distance traveled in (1) first 3 sec (2) 9th second of its motion.​

Answer 1

Step-by-step explanation:

Answer :-

Distance travelled by the body :-

→ In first 3 sec is 27 m .

→ 9th second of it's motion is 16 m .

Explanation :-

Firstly, let's calculate the acceleration and initial velocity of the body.

For 3rd second :-

⇒ 10 = u + a/2[2 × 3 - 1]

⇒ 10 = u + 5a/2

⇒ 10 = u + 2.5a -----(1)

For 5th second :-

⇒ 12 = u + a/2[2 × 5 - 1]

⇒ 12 = u + 9a/2

⇒ 12 = u + 4.5a -----(2)

Subtracting eq.2 from eq.1 :-

⇒ 10 - 12 = u + 2.5a - u - 4.5a

⇒ - 2 = - 2a

⇒ a = 1 m/s²

Putting value of 'a' in eq.1 :-

⇒ 10 = u + 2.5(1)

⇒ u = 10 - 2.5

⇒ u = 7.5 m/s

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By using the 2nd equation of motion :-

s = ut + ½at²

⇒ s = 7.5(3) + ½ × 1 × (3)²

⇒ s = 22.5 + 4.5

⇒ s = 27 m

Now, from the equation of distance covered in the 'nth' second :-

sₙₜₕ = u + a/2(2n - 1)

⇒ s₉ₜₕ = 7.5 + 1/2[2 × 9 - 1]

⇒ s₉ₜₕ = 7.5 + 0.5[17]

⇒ s₉ₜₕ = 7.5 + 8.5

⇒ s₉ₜₕ = 16 m

Answer 2

Let me try to explain. uniform acceleration means that (v) of the object increases at a constant rate. so (a) is a constant. the formula is d=vt + 1/2at^2. so if the object travels 5m from 0 to 3sec we apply d=0.5a3^2 where vt = 0 so d=0.5*9*a

5m=9/2a or a =10/9m/s*s. we olny need that to calculate the distance traveled in 6 sec since a is a constant. so in 6 sec d=1/2at^2 d= 0.5*10/9*6^2 = 180/9=20m.

so your description that from the third to the fifth sec it does just 4m is wrong unless the a is not uniform which by your question it is. the distance between the 3rd and 5 second is d=vt + 1/2at^2 where v=at = 10/9*3= 30/9 m/s in t=3sec now

d=30/9*2 + 0.5*10/9*2^2= 60/9 +10/4= 240+90/36 = 330/36 m = 9.16m.

i hope my memories haven`t let me down yet cause i have 10+ years that i don`t practice