JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 1)

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 :
10 m
20 m
30 m
40 m

Explanation

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:

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

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

$$= 30 \text{ m}$$

Comments (0)

Advertisement