Animated Solution for Physics - Kinematics: The position vector of particle changes with time according to the relation $\mathbf{r}(t)=15t^2\hat{\mathbf{i}}+(4-20t^2)\hat{\mathbf{j}}$. What is the magnitude of the acceleration (in ${\text{ms}}^{-2}$) at $t=1$?

In this problem, we are given the position vector of a particle as a function of time, and we need to find the magnitude of its acceleration at a specific instant, $t=1$ second. This is a classic kinematics problem that tests our understanding of the relationship between position, velocity, and acceleration using calculus.

The position vector is given by:

This vector tells us exactly where the particle is on the 2D plane at any given time $t$. The $\hat{\mathbf{i}}$ component represents the horizontal position, and the $\hat{\mathbf{j}}$ component represents the vertical position. Notice that both components depend on $t^2$, which hints that the motion is not uniform.

To find the acceleration, we must first find the velocity. Velocity is the rate of change of position with respect to time. Mathematically, it is the first derivative of the position vector:

Let's differentiate the position vector term by term. The derivative of $15t^2$ is $30t$. For the $\hat{\mathbf{j}}$ component, the derivative of the constant $4$ is $0$, and the derivative of $-20t^2$ is $-40t$. Putting it all together, we get the velocity vector:

Now that we have the velocity, we can find the acceleration. Acceleration is the rate of change of velocity with respect to time, which means we need to take the derivative of the velocity vector:

Differentiating $30t$ gives us $30$, and differentiating $-40t$ gives us $-40$. Therefore, the acceleration vector is:

A crucial observation here: The time variable $t$ has completely disappeared from our acceleration equation! This means that the acceleration is constant. It does not matter whether we are looking at $t=1$ second, $t=5$ seconds, or $t=100$ seconds; the acceleration vector remains exactly the same.

The question asks for the magnitude of the acceleration. Since the $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ components are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant vector:

Substituting our components $a_x=30$ and $a_y=-40$ into the formula:

Thus, the magnitude of the acceleration is $50{\text{ ms}}^{-2}$. The correct option is (a).