- Home
- Standard 11
- Physics
3-2.Motion in Plane
medium
A bus is moving on a straight road towards north with a uniform speed of $50\; km / hour$ then it turns left through $90^{\circ} .$ If the speed remains unchanged after turning, the increase in the velocity of bus in the turning process is
A$50 \;km / hr$ along west
B$0$
C$70.7 \;km / hr$ along south-west direction
D$70.7\; km / hr$ along north-west direction
(AIPMT-1989)
Solution
$v_{1}=50 km / hr$ due north
$v_{2}=50 km / hr \text { due west }$
$-v_{1}=50 km / hr \text { due south }$
Magnitude of change in velocity
$=\left|\vec{v}_{2}-\vec{v}_{1}\right|=\left|\vec{v}_{2}+\left(-\vec{v}_{1}\right)\right|$
$=\sqrt{v_{2}^{2}+\left(-v_{1}\right)^{2}}$
$=\sqrt{(50)^{2}+(50)^{2}}=70.7 km / hr$
$\vec{v}=70.7 km / hr$ along south-west direction
$v_{2}=50 km / hr \text { due west }$
$-v_{1}=50 km / hr \text { due south }$
Magnitude of change in velocity
$=\left|\vec{v}_{2}-\vec{v}_{1}\right|=\left|\vec{v}_{2}+\left(-\vec{v}_{1}\right)\right|$
$=\sqrt{v_{2}^{2}+\left(-v_{1}\right)^{2}}$
$=\sqrt{(50)^{2}+(50)^{2}}=70.7 km / hr$
$\vec{v}=70.7 km / hr$ along south-west direction
Standard 11
Physics
