The rate of change of displacement of a body with respect to its surrounding, is known
A. Velocity
B. Acceleration
C. Speed
D. None of these
Answer: Option A
Solution (By Examveda Team)
Displacement refers to the change in position of a body in a particular direction.The rate of change of displacement with respect to time is defined as velocity.
Velocity is a vector quantity, meaning it has both magnitude and direction.
In contrast, speed is the rate of change of distance (not displacement) and is a scalar quantity.
Acceleration is the rate of change of velocity, not displacement.
Therefore, the correct term for the rate of change of displacement is Velocity.
This Question Belongs to Civil Engineering >> Applied Mechanics And Graphic Statics
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Comments (1)
In case of S.H.M. the period of oscillation (T), is given by
A. $${\text{T}} = \frac{{2\omega }}{{{\pi ^2}}}$$
B. $${\text{T}} = \frac{{2\pi }}{\omega }$$
C. $${\text{T}} = \frac{2}{\omega }$$
D. $${\text{T}} = \frac{\pi }{{2\omega }}$$
The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be
A. 0.1 rad/sec
B. 1 rad/sec
C. 10 rad/sec
D. 100 rad/sec
A body is said to move with Simple Harmonic Motion if its acceleration, is
A. Always directed away from the centre, the point of reference
B. Proportional to the square of the distance from the point of reference
C. Proportional to the distance from the point of reference and directed towards it
D. Inversely proportion to the distance from the point of reference
The resultant of two forces P and Q acting at an angle $$\theta $$, is
A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$
B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$
C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$
D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$
E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
a;velocity