Let v and a denote the velocity and accelerat | vedclass

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

Question

(A) The speed of a particle is the magnitude of its velocity vector $v$. The rate of change of speed is given by the tangential component of accelerat

Vedclass · Accepted Answer

The speed of the particle decreases when $v \cdot a < 0$.

Vedclass · Answer

(A) The speed of a particle is the magnitude of its velocity vector $v$. The rate of change of speed is given by the tangential component of acceleration,$a_t = \frac{dv}{dt} = \frac{a \cdot v}{|v|}$.$1$. If $v \cdot a > 0$,the angle between velocity and acceleration is acute $(0^\circ \le 	heta $2$. If $v \cdot a $3$. The cross product $v 	imes a$ relates to the change in direction of velocity (centripetal acceleration),not the change in speed in one-dimensional motion. Thus,option $(a)$ is correct.

Let $v$ and $a$ denote the velocity and acceleration respectively of a particle in one-dimensional motion. Which of the following statements is correct?

Similar Questions

If the velocity of a particle is $(10 + 2t^2) \ m/s$,then the average acceleration of the particle between $2 \ s$ and $5 \ s$ is..........$m/s^2$.

Explain the acceleration.

If the speed of a moving object decreases,what are the directions of its velocity and acceleration?

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \,ms^{-1}$ at $x=0$, then its velocity at $x=1.4 \,m$ is (in $ms^{-1}$)

$A$ body starts from the origin and moves along the $x$-axis such that its position at any instant is $x = 4t^2 - 12t$,where $t$ is in seconds and $x$ is in meters. What is the acceleration of the particle in $m/s^2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module