A train starting from rest accelerates unifor | vedclass

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(D) The distance covered is equal to the area under the velocity-time graph.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 3000\,m.$$\f

Vedclass · Accepted Answer

The maximum velocity is $10\,m/s$

Vedclass · Answer

(D) The distance covered is equal to the area under the velocity-time graph.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 3000\,m.$$\frac{1}{2} 	imes 300\,s 	imes V_{\max} = 3000\,m.$$150 	imes V_{\max} = 3000 \implies V_{\max} = 20\,m/s.$Acceleration $a_1 = \frac{V_{\max} - 0}{100} = \frac{20}{100} = 0.2\,m/s^2.$Retardation $a_2 = \frac{0 - V_{\max}}{200} = -\frac{20}{200} = -0.1\,m/s^2$ (magnitude is $0.1\,m/s^2$).Since $V_{\max} = 20\,m/s,$ option $D$ is incorrect.

$A$ train starting from rest accelerates uniformly for $100\,s,$ then comes to a stop with a uniform retardation in the next $200\,s.$ During the motion,it covers a distance of $3\,km.$ Choose the wrong option.

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