A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s, the distance travelled by the toy in the next t s will be :

Motion in A Straight Line PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q44 with Solution

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Correct answer: C

Given:
- The toy starts from rest, so initial velocity $u=0$ .
- It moves with constant acceleration $a$ .
- Distance travelled in first $t$ seconds is $10\,\text{m}$ .
Distance in first $t$ seconds: Using $s=ut+\frac{1}{2}at^2$ Since $u=0$ , $s_1=\frac{1}{2}at^2$ Given $s_1=10$ , $\frac{1}{2}at^2=10$ $at^2=20$
Distance in first $2t$ seconds: $s(2t)=u(2t)+\frac{1}{2}a(2t)^2$ Since $u=0$ , $s(2t)=\frac{1}{2}a\cdot 4t^2=2at^2$ Using $at^2=20$ , $s(2t)=2\times 20=40\,\text{m}$
Distance in the next $t$ seconds: This means distance from time $t$ to time $2t$ : $s_2=s(2t)-s(t)=40-10=30\,\text{m}$
Option check:
- A: $10\,\text{m}$ ❌
- B: $20\,\text{m}$ ❌
- C: $30\,\text{m}$ ✅
- D: $40\,\text{m}$ ❌

Therefore, the correct answer is C.