Magnitude of acceleration and direction

To find the magnitude and direction of acceleration, we need to determine the second derivative of the position vector ( \mathbf{r}(t) ) with respect to time ( t ).

Given: \(\mathbf{r}(t) = (3.2t + 1.5t^2) \mathbf{i} + (1.7t - 2.0t^2) \mathbf{j}\)

First, let’s find the velocity ( \mathbf{v}(t) ) by taking the first derivative of ( \mathbf{r}(t) ):

\(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\)

For the \(\mathbf{i}\) component: \(v_i(t) = \frac{d}{dt}(3.2t + 1.5t^2) = 3.2 + 3.0t\)

For the \(\mathbf{j}\) component: \(v_j(t) = \frac{d}{dt}(1.7t - 2.0t^2) = 1.7 - 4.0t\)

So, the velocity vector is: \(\mathbf{v}(t) = (3.2 + 3.0t) \mathbf{i} + (1.7 - 4.0t) \mathbf{j}\)

Next, let’s find the acceleration ( \mathbf{a}(t) ) by taking the second derivative of ( \mathbf{r}(t) ):

\(\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}\)

For the \(\mathbf{i}\) component: \(a_i(t) = \frac{d}{dt}(3.2 + 3.0t) = 3.0\)

For the \(\mathbf{j}\) component: \(a_j(t) = \frac{d}{dt}(1.7 - 4.0t) = -4.0\)

So, the acceleration vector is: \(\mathbf{a}(t) = 3.0 \mathbf{i} - 4.0 \mathbf{j}\)

To find the magnitude of the acceleration: \(|\mathbf{a}(t)| = \sqrt{(3.0)^2 + (-4.0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0 , \text{m/s}^2\)

To find the direction of the acceleration, we calculate the angle ( \theta ) with respect to the positive ( \mathbf{i} ) axis: \(\theta = \tan^{-1}\left(\frac{a_j}{a_i}\right) = \tan^{-1}\left(\frac{-4.0}{3.0}\right)\)

\(\theta = \tan^{-1}(-\frac{4}{3}) \approx -53.13^\circ\)

So, the magnitude of the acceleration is \(5.0 , \text{m/s}^2\) and the direction is approximately \(-53.13^\circ\) with respect to the positive \(\mathbf{i}\) axis.

Is anyone not seeing the math expressions correctly?

Fixed it

Hope @bolbol doesn't mind!