Kinematics

JEE Main level, single correct, medium

25 of 308

Question 25 of 308

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m10m in tsts, the distance travelled by the toy in the next tsts will be:

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Answer: (C) 30m30m

Initial velocity of the particle is zero.
Applying equation of motion, we get 10=12×a×t210=\dfrac{1}{2}\times a\times {t}^{2}
Let distance covered in next tsts is xx.
Then, 10+x=12×a×(2t)210+x=\dfrac{1}{2}\times a\times {\left(2t\right)}^{2}
⇒10+x=12at2×4\Rightarrow 10+x=\dfrac{1}{2}a{t}^{2}\times 4
⇒10+x=10×4\Rightarrow 10+x=10\times 4
⇒x=30m\Rightarrow x=30m

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