A particle starts from rest. Its acceleration | vedclass

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(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

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 $m \ s^{-1}$)

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