Question:
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 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
Updated On: May 5, 2024
- 70.7 km/hr along south-west direction
- zero
- 50 km/hr along west
- 70.7 km/hr along north-west direction
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The Correct Option is A
Solution and Explanation
$ v_1 $ 50 km/hr due north
$ v_2$ 50 km/hr due west
- $ v_1$ 50 km/hr due south
Magnitude of change in velocity
= $ | \overrightarrow{v_2} - \overrightarrow{v_1} | = | \overrightarrow{v_2} + ( - \overrightarrow{v_1}) | $
= $ \sqrt{ v_2^2 + ( - v_1)^2 }$
= $ \sqrt{ ( 50)^2 + (50)^2 } $
= 70.7 km/hr along south-west direction
$ v_2$ 50 km/hr due west
- $ v_1$ 50 km/hr due south
Magnitude of change in velocity
= $ | \overrightarrow{v_2} - \overrightarrow{v_1} | = | \overrightarrow{v_2} + ( - \overrightarrow{v_1}) | $
= $ \sqrt{ v_2^2 + ( - v_1)^2 }$
= $ \sqrt{ ( 50)^2 + (50)^2 } $
= 70.7 km/hr along south-west direction
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration