Kerala KEAM 5 June 2024 Solved Paper

The position vector of the particle is given by: $$\mathbf{r}(t) = 3 \hat{i} + 2 t^2 \hat{j}$$To find the velocity, we differentiate the position vector with respect to time:$$\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)$$Differentiating each component:$$\mathbf{v}(t) = \frac{d}{dt} (3 \hat{i}) + \frac{d}{dt} (2 t^2 \hat{j})$$$$\mathbf{v}(t) = 0 \hat{i} + 4 t \hat{j}$$Thus, the velocity vector is:$$\mathbf{v}(t) = 4 t \hat{j}$$At $t = 5$ seconds:$$\mathbf{v}(5) = 4 \times 5 \hat{j} = 20 \hat{j} \, m/s$$The magnitude of the velocity is:$$| \mathbf{v}(5) | = 20 \, m/s$$Thus, the magnitude of the velocity at $t = 5$ seconds is $20 \, ms^{-1}$.$$\boxed{20}$$ Quick Tip: The velocity is the derivative of the position vector with respect to time. For each component, differentiate and find the magnitude of the velocity vector.