Answer:
Reaction time, \[t=0.20s\]
initial speed of the car, u = 54 km/hr
\[=54\times \frac{5}{18}=15m/s\]
deceleration, \[a=-6.0m{{s}^{-2}}.\]
distance travelled during the reaction time,
\[{{S}_{1}}=u\times {{t}_{1}}=15\times 0.2=3m\]
After applying the breaks, the car travelled the distance,
\[{{S}_{2}}=\frac{{{v}^{2}}-{{u}^{2}}}{2a}=\frac{0-{{(15)}^{2}}}{-2\times 6}=\frac{225}{12}=18.75m\]
total distance travelled by the can after the sees the need to put the brakes on
\[S={{S}_{1}}+{{S}_{2}}=3+18.75=21.75m\approx 22m\]
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