Question

A particle experiences a constant acceleration for 20 seconds after starting from rest. If 8t travels, a distance S1 in the first 10 seconds and distance S2 in next 10 seconds then A) S1=S2 B)S2=2S1 C)S2=3S1 D)S2=4S1​

Answer 1

➩ Given :-

• Initial velocity = 0
• Distance travelled in first 10s = S1
• Distance travelled in next 10s = S2

➩ To Find :-

• Relation between S1 and S2

➩ Solution :-

✦ Let a be the constant acceleration of the particle. Then,

⇒ s = ut+ 1/2 at ^2

⇒S1 = 0+ 1/2 × a× (10)^2=50a

⇒S2=[0+ 1/2 a(20) ^2] - 50a = 150a

⇒S2 = 3 s1

▶Alternatively Let a be constant acceleration and s= ut + 1/2 at ^2

▶ Then, s1 = 0 + 1/2 × a × 100 = 50a

▶Velocity after 10 sec will be

⇒ v = 0 + 10a

⇒S2 =10a×10+ 21a×100

⇒S2 =150 a

⇒S2 = 3s1

▶Now Let a be constant acceleration,

using s=ut+ 1/2 at ^2

✪ Distance covered in first 10 seconds: ⇒S1 =0+ 1/2 × a × 100 = 50a

▶Velocity after 10 sec is v=0+10a.

✪ Distance covered in next 10 seconds:

⇒S2 = 10a × 10 + 21 a×100 = 150a

⇒S2 = 3S1

Therefore Option C is the correct answer!

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Answer 2

Given :

Initial velocity = zero

Distance travelled in first 10s = S₁

Distance travelled in next 10s = S₂

To Find :

Relation between S₁ and S₂.

Solution :

❖ Since acceleration is said to be constant, equation of kinematics can be easily applied to solve this question$.$

Second equation of kinematics :

• d = ut + 1/2 at²

» d denotes distance

» u denotes initial velocity

» t denotes time

» a denotes acceleration

A] Distance travelled in first 10 s :

➙ S₁ = (0 × 10) + 1/2 a (10)²

➙ S₁ = 100a/2

➙ S₁ = 50a .......... [i]

B] Total distance travelled in 20 s :

➙ S = (0 × 20) + 1/2 a (20)²

➙ S = 400a/2

➙ S = 200a .......... [ii]

C] Distance travelled in next 10 s :

➙ S = S₁ + S₂

➙ 200a = 50a + S₂

➙ S₂ = 150a ......... [iii]

Taking ratio of [i] and [iii], we get

➠ S₁ / S₂ = 50a / 150a

➠ S₁ / S₂ = 1 / 3

➠ S₂ = 3S₁