IGNOU PHE 4 2018 - The position vector of an object is:r = (1+t)3/2i + (1– t) 3

Question

The position vector of an object is:

r = (1+t)^3/2i + (1– t) ^3/2j + t²k

Determine the angle between its velocity vector and acceleration vector at t = 0.

25 Jan 2018

Answer :

Word Count : 393

Let's solve this step by step, manually. We are given the position vector: $$ \mathbf{r}(t) = (1+t)^{3/2}\mathbf{i} + (1-t)^{3/2}\mathbf{j} + t^2 \mathbf{k} $$ We are asked to find the angle between the velocity and acceleration vectors at $t = 0$. --- ### Step 1: Find the velocity vector Velocity is the derivative of position: $$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} $$ Compute each component: 1. $i$-component: $\frac{d}{dt} (1+t)^{3/2} = \frac{3}{2} (1+t)^{1/2}$ 2. $j$-component: $\frac{d}{dt} (1-t)^{3/2} = \frac{3}{2} (1-t)^{1/2} \cdot (-1) = -\frac{3}{2} (1-t)^{1/2}$ 3. $k$-component: $\frac{d}{dt} t^2 =

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