Match the following columns.Column IColumn II | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $\frac{dv}{dt}$ is the definition of acceleration, which is the rate of change of velocity vector. Thus, $(A \rightarrow p)$.$(B)$ $\frac{d|v|}{dt

Vedclass · Accepted Answer

$(A \rightarrow p, B \rightarrow q, C \rightarrow r, D \rightarrow s)$

Vedclass · Answer

(A) $\frac{dv}{dt}$ is the definition of acceleration, which is the rate of change of velocity vector. Thus, $(A \rightarrow p)$.$(B)$ $\frac{d|v|}{dt}$ represents the rate of change of the magnitude of velocity, which is the tangential acceleration or the rate of change of speed. Thus, $(B \rightarrow q)$.$(C)$ $\frac{dr}{dt}$ is the definition of velocity, which is the rate of change of the position vector. Thus, $(C \rightarrow r)$.$(D)$ $\left|\frac{dr}{dt}\right|$ is the magnitude of the velocity vector, which is the speed of the particle. Thus, $(D \rightarrow s)$.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d\|v\|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left\|\frac{dr}{dt}\right\|$	$(s)$ Magnitude of velocity

Match the following columns.
Column $I$ Column $II$
$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$ $(s)$ Magnitude of velocity

Similar Questions

$A$ particle starts from rest. Its acceleration $a$ versus time $t$ is shown in the figure. The maximum speed of the particle will be (in $m/s$)

The motion of a particle along a straight line is described by the function $x = (2t - 3)^2$,where $x$ is in metres and $t$ is in seconds. The acceleration of the particle at $t = 2 \,s$ is (in $\,m/s^2$)

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the graph. If the velocity of the particle is $0.8 \ m/s$ at $x = 0$,find the velocity of the particle at $x = 1.4 \ m$ (in $m/s$).

The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle will be zero at time $t$ equal to

The displacement is given by $x = 2t^2 + t + 5$. The acceleration at $t = 2 \ s$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match the following columns.Column $I$Column $II$$(A)$ $\frac{dv}{dt}$$(p)$ Acceleration$(B)$ $\frac{d|v|}{dt}$$(q)$ Magnitude of acceleration$(C)$ $\frac{dr}{dt}$$(r)$ Velocity$(D)$ $\left|\frac{dr}{dt}\right|$$(s)$ Magnitude of velocity