A) 19 m
B) 9 m
C) 10 m
D) 28 m
Correct Answer: A
Solution :
Before deceleration the car has covered the distance \[{{s}_{1}}=ut\] Given: u= 36 km/h \[u=\frac{36\times 5}{18}=10\,m/s\] \[t=0.9\,s\] \[{{s}_{1}}=10\times 0.9=9\,m\] Again the time taken in travelling by the car after deceleration \[v=u+gt\] \[\left( \begin{align} & \because \,v=0 \\ & a=-5\,m/{{s}^{2}} \\ \end{align} \right)\] \[0=10-5{{t}_{1}}\] \[{{t}_{1}}=\frac{10}{5}=2s\] Now, the distance travelled by the car after deceleration is given by \[{{s}_{2}}=u{{t}_{1}}+\frac{1}{2}a{{t}^{2}}\] \[=10\times 2-\frac{1}{2}\times 5\times {{(2)}^{2}}\] \[20-10=10\,m\] Hence, total distance travelled by car is given by = 9m + 10 m = 19m
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