A particle starts from rest. Its acceleration | vedclass

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(B) The area under the acceleration-time graph gives the change in velocity of the particle.Since the particle starts from rest,its initial velocity $

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) The area under the acceleration-time graph gives the change in velocity of the particle.Since the particle starts from rest,its initial velocity $u = 0$.The maximum velocity is achieved at $t = 11 \, s$,where the acceleration becomes zero.Thus,the maximum velocity $v_{\max}$ is equal to the area of the triangle $OAB$ in the given graph.$v_{\max} = 	ext{Area of } \Delta OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$v_{\max} = \frac{1}{2} 	imes 11 \, s 	imes 10 \, m/s^2 = 55 \, m/s$.Therefore,the maximum speed of the particle is $55 \, m/s$.

$A$ 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.....$m/s$.

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