Question

A man is running due north at 2m/s turns east and runs at the same speed. What is the change in velocity?

Answer 1

Explanations & Calculations

• Change of velocity is given by $\small \Delta \vec{v}=\vec{v_{final}}-\vec{v_{initial}}$. All you need to know in calculating this is treating them properly as they are vectors.
• Each velocity has its own magnitude & direction. For this case, magnitude is the same in both stuations.
• The change in velocity also has a magnitude and a direction.

• We are given,

$\qquad\qquad \begin{aligned} \small \vec{v_{final}}&=\small 2[\to E]\,ms^{-1}\\\\ \small \vec{v_{initial}} &=\small 2[\uparrow N]\,ms^{-1} \end{aligned}$

• Then,

$\qquad\qquad \begin{aligned} \small \Delta v&=\small 2[\to E]-2[\uparrow N] \end{aligned}$

• Since we know only to add velocities, we need to rearrange the above relationship.

$\qquad\qquad \begin{aligned} \small \Delta v&=\small 2[\to E]+2[-\uparrow N]\\ &=\small 2[\to E]+2[\downarrow S] \end{aligned}$

• Then if you place them in a velocity triangle, they will form a right angled one with the resultant/change is given by its hypotenuse both in direction & magnitude.
• Therefore,

$\qquad\qquad \begin{aligned} \small |\vec{\Delta v}|&=\small \sqrt{2^2+2^2}\\ &=\small \bold{2\sqrt2} \,ms^{-1}\\\\ \small \theta&=\small \tan^{-1}\bigg(\frac{2\downarrow}{2\to}\bigg)\\ &=\small 45^0\\\\ \small \vec{\Delta v}&=\small \bold{2\sqrt2\,ms^{-1}}\text{due southeast} \end{aligned}$