QJEEM22Phc83c: JEE Main 2022 Physics - Class 11 JEEM22Phc83c Solution: A small toy starts moving from the position of rest under a…

A small toy begins to move from a standstill with a constant acceleration. It covers a distance of 10 meters in t seconds. We need to determine the distance the toy will travel in the subsequent t seconds.

First, we know that the initial distance traveled is given by the formula for constant acceleration starting from rest:

$\frac{1}{2} a t^2 = 10 \text{ m}$

Next, we calculate the total distance traveled after 2t seconds:

$\frac{1}{2} a (2t)^2 = \frac{1}{2} a \cdot 4t^2 = 2 a t^2$

Since we know from the initial condition that $\frac{1}{2} a t^2 = 10 \text{ m}$ , multiplying it by 4 gives

$2 a t^2 = 40 \text{ m}$

Thus, the additional distance traveled in the next t seconds is:

$\text{Total distance after 2t seconds} - \text{Distance already traveled in t seconds}$

$= 40 \text{ m} - 10 \text{ m}$

$= 30 \text{ m}$