The reaction time for an automobile driver is $$0.7$$ $$sec$$. If the automobile can be decelerated at $$5$$ $$m/s2$$ calculate the total distance travelled in coming to stop from an initial velocity of $$8.33$$ $$m/s$$ after a signal is observed.

v $$=$$ u $$+$$ $$at0$$ $$=$$ $$8.33$$ $$-$$ $$5$$ x t⇒t $$=$$ $$8.33/5displacement$$ $$s=ut+u22aArea$$ $$=8.33$$×$$0.7+12$$×$$8.3325=$$ $$12.77$$ m

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The reaction time for an automobile driver is 0.7 sec. If the automobile can be decelerated at 5 m/s² calculate the total distance travelled in coming to stop from an initial velocity of 8.33 m/s after a signal is observed.

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12.77 m

14.82 m

16.83 m

19.65 m

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detailed solution

Correct option is A

v = u + at
0 = 8.33 - 5 x t $\Rightarrow$ t = 8.33/5
displacement $s = ut + \frac{u^{2}}{2 a}$
Area $= 8.33 \times 0.7 + \frac{1}{2} \times \frac{{8.33}^{2}}{5}$
= 12.77 m

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The reaction time for an automobile driver is 0.7 sec. If the automobile can be decelerated at 5 m/s2 calculate the total distance travelled in coming to stop from an initial velocity of 8.33 m/s after a signal is observed.

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The reaction time for an automobile driver is 0.7 sec. If the automobile can be decelerated at 5 m/s² calculate the total distance travelled in coming to stop from an initial velocity of 8.33 m/s after a signal is observed.