CBSE-XI-Physics
03: Rest and Motion: Kinematics
- #1A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about
-(a) 50 km/h towards west
(b) 70 km/h towards south-west
(c) 70 km/h towards north-west
(d) zero.digAnsr: bAns : (b) 70 km/h towards south-west
Final velocity, `` \stackrel{\to }{{V}_{f}}=-50\stackrel{‸}{i}\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
Initial velocity, `` \stackrel{\to }{{V}_{i}}=50\stackrel{‸}{j}\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
Change in velocity, `` ∆\stackrel{\to }{V}=\stackrel{\to }{{V}_{f}}-\stackrel{\to }{{V}_{i}}``
`` \left|∆V\right|=\sqrt{{50}^{2}+{50}^{2}+2\times 50\times 50\,\mathrm{\,cos \,}\left(90°\right)}=70\,\mathrm{\,km \,}/\,\mathrm{\,h \,}``
It is towards southwest, as shown in the figure.
