Answer:
\[{{t}_{reaction}}=0.5s\,\,u=72\,kmph\]
\[=72\times \frac{5}{18}=20m/s\]
\[{{S}^{1}}=ut=0.5\times 20=10m\] Now distance left
\[=50-10=40m,\,\,v=0,\,\,a=?\]
using \[{{v}^{2}}-{{u}^{2}}=2as\]
\[\Rightarrow {{0}^{2}}-{{(20)}^{2}}=2\times a\times 40\]
\[\Rightarrow -400=80a\Rightarrow a=-5m/{{s}^{2}}\]
\[\therefore \]The minimum deceleration produced by the brakes of the car is\[5m/{{s}^{2}}.\]
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