Question

A particle experiences constant acceleration for 20 seconds after starting from rest. if it travels a distance s1 in the first 10 seconds and distance s2 in the next 10 seconds of its motion define relation between s¹ and s²

Answer 1

Given :

▪ Initial velocity = zero

▪ Total time of journey = 20s

▪ Distance covered in first 10s = S¹

▪ Distance covered in last 10s = S²

To FinD :

▪ Relation b/w S¹ and S²

SoLuTioN :

Let total distance covered by body in 20s be S.

Initial velocity (u) = zero

✴ Total distance covered in 20s :

☞ S = ut + (1/2)at²

☞ S = (0×20) + (1/2)a(20)²

☞ S = 0 + (1/2)a(400)

☞ S = 200a ...... (i)

✴ Distance covered in first 10s :

↗ S¹ = ut + (1/2)at²

↗ S¹ = (0×10) + (1/2)a(10)²

↗ S¹ = 0 + (1/2)a(100)

↗ S¹ = 50a ...... (ii)

✴ Distance covered in last 10s :

⤵ S² = S - S¹

⤵ S² = 200a - 50a

⤵ S² = 150a ...... (iii)

☣ Relation between S¹ and S² :

➡ S¹ : S² = 50a : 150a

➡ S¹ : S² = 1 : 3

Answer 2

Given:

• Total time = 20s
• Distance in first 10s = S¹
• Distance in next 10s = S²

To find:

• Relationship between S¹ & S²

Solution:

Here , initial velocity = 0 m/s because the particle starts from rest.

Total distance = S¹ + S²

Let total distance be S

S = S¹ + S²

S² = S - S¹

Firstly finding distance ( S¹ )

Using

S = ut + 1/2 at²

S¹ = 0(10) + 1/2(a)(10)²

S¹ = 1/2 × a × 100

S¹ = 50a

Similarly

S = 0 × 20 + 1/2 × a × 20²

S = 1/2 × a × 400

S = 200a

Now , S²

S² = S - S¹

S² = 200a - 50a

S² = 150a

Now , relationship between S¹ & S²

S¹ : S²

♦ 50a : 150a

♦ 1 : 3

Hence ,. relationship between S¹ & S² = 1 : 3