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(B) The acceleration is given by the slope of the speed-time graph.For part $AB$: Slope $= \frac{20 - 0}{0.75 - 0} = \frac{20}{0.75} \approx 26.67 \;

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) The acceleration is given by the slope of the speed-time graph.For part $AB$: Slope $= \frac{20 - 0}{0.75 - 0} = \frac{20}{0.75} \approx 26.67 \; km\, h^{-2}$.For part $BC$: Slope $= 0$ (constant speed).For part $CD$: Slope $= \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160 \; km\, h^{-2}$.For part $DE$: Slope $= \frac{0 - 60}{2.00 - 1.25} = \frac{-60}{0.75} = -80 \; km\, h^{-2}$ (deceleration).Comparing these values,the maximum acceleration is $160 \; km\, h^{-2}$,which corresponds to the segment $CD$.

$A$ train moves from one station to another in $2$ hours. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is ............. $km\, h^{-2}$.

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