Question

A bdy moving with speed of 30km/hr towards east turns sharply towards north , the magnitude of change of velocity is

Answer 1

Answer:

a car Will goes to 40 km/hr

Answer 2

Answer:

30√2 km/hr

Explanation:

Let the east be +x axis and north be +y axis

Intital velocity vi = (30î)km/hr

Final velocity vf = (30j)km/hr

Change in velocity = vf - vi

= 30j - 30î

magnitude |vf - vi| = | (-30i + 30j) |

= √ 30² + 30² + 2(30)²cos90°

= √ 2×30²

= 30√2 km/hr