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 particle starts from rest,so its initial velocity $u = 0 \ m \ s^{-1}$.In an acceleration-time $(a-t)$ graph,the change in velocity $(\Delta v

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) The particle starts from rest,so its initial velocity $u = 0 \ m \ s^{-1}$.In an acceleration-time $(a-t)$ graph,the change in velocity $(\Delta v)$ is equal to the area under the curve.Since the acceleration is positive throughout the interval from $t = 0 \ s$ to $t = 15 \ s$,the velocity of the particle increases continuously.Therefore,the maximum speed is attained at $t = 15 \ s$.The area under the $a-t$ graph is the area of the right-angled triangle with base $15 \ s$ and height $10 \ m \ s^{-2}$.$\Delta v = 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$\Delta v = \frac{1}{2} 	imes 15 \ s 	imes 10 \ m \ s^{-2} = 75 \ m \ s^{-1}$.Since $v_{max} = u + \Delta v = 0 + 75 \ m \ s^{-1} = 75 \ m \ s^{-1}$.

$A$ particle starts from rest. Its acceleration $(a)$ versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be (in $m \ s^{-1}$)

Similar Questions

What is retardation?

The displacement of a particle,moving in a straight line,is given by $s = 2t^2 + 2t + 4$,where $s$ is in metres and $t$ is in seconds. The acceleration of the particle is ........ $m/s^2$.

The acceleration-time graph for a particle moving along the $x$-axis is shown in the figure. If the initial velocity of the particle is $-5 \, m/s$, the velocity at $t = 8 \, s$ is ....... $m/s$.

What can be the angle between velocity and acceleration for the motion on a straight line? Explain with an example.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module