A particle starts from rest. Its acceleration | vedclass

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

Question

(B) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.Given that the particle starts from rest,the i

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.Given that the particle starts from rest,the initial velocity $u = 0$.The area under the $a-t$ graph is a triangle with base $b = 10 \ s$ and height $h = 8 \ m/s^2$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Area $= \frac{1}{2} 	imes 10 	imes 8 = 40 \ m/s$.Since $\Delta v = v - u = 	ext{Area}$,and $u = 0$,we have $v = 40 \ m/s$.Thus,the maximum speed of the particle is $40 \ m/s$.

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

Similar Questions

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

The position $x$ of a particle varies with time as $x = at^2 - bt^3$,where $a$ and $b$ are constants. The acceleration of the particle will be zero at:

Define acceleration,average acceleration,and instantaneous acceleration.

The velocity of a body depends on time according to the equation $v = 20 + 0.1t^2$. The body is undergoing

$A$ body is moving with variable acceleration $a$ along a straight line. The average acceleration of the body in the time interval $t_1$ to $t_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module