Question

A motar car is going due north at a speed of 50 km per hour it's make a 90 degree left turn without changing the speed the changing in the velocity of the car is about​

Answer 1

$\bold{Initial\:velocity}$ of the car is, 

$v_{1}$ = ${50\:km/h}$ towards north.

$\bold{Final\:velocity}$ of the car is, 

$v_{2}$ = ${50\:km/h}$ towards west.

Change in $\bold{velocity,}$

∆v = ${v_{2}-v_{1}}$

$\bold{Magnitude}$ of the change is,

|∆v|$= (v_{2}^{2} + v_{1}^{2} - 2v_{1}v_{2} \: cos90 ^{°} ) ^{ \frac{1}{2} }$

|∆v|${ = 70.71\:km/h}$

The direction of this change is φ w.r.t the initial direction,

$tan^{ - 1} (\frac{v_{2}}{ - v_{1}} )$

$= tan ^{ - 1} ( - 1)$

${ = 135°}$