SSC JE Civil Engineering 29-Oct-2020 Shift 1 Paper

Question : 142 of 200

Marks: +1, -0

Two cars are moving in the opposite direction with speeds 25 m/s and 20 m/s respectively. If total reaction time is 2.5 seconds and design coefficient of friction is 0.35, then what must be the minimum sight distance to avoid head on collision? (Take

g = 10 \m\/\s^2

)

112.5 m
34.07 m
225 m
258.93 m

Validate

Solution:

Consider two cars are approaching each other as shown above, To avoid any head-on collision,

\SSD = \SD_1 + \SD_2

For car A :

\f=0.35, t_{r}=2.5,

\V_{\A}=25 {\m} \/ {\s}

∴\SSD_{1}=\V_{\A} .t_{r}+{(\V_{\A})^{2}}/{2 \gf}

⇒\SSD_{1}

=25 ×2.5+{25^{2}}/{2 ×10 ×0.35}

=151.78 {\m}

For car B :

{\V}_{{\B}}=20 {\m} \/ {\s}

{\f}=0.35, {\t}_{r}=2.5

∴\SSD_{2}=\V_{\B} .t_{r}+{(\V_{\B})^{2}}/{2 \gf}

=20 ×2.5+{(20)^{2}}/{2 ×10 ×0.35}

=107.14 {\m}

∴ Total sight distance

= \SSD_1 + \SSD_2

= 151.78 + 107.14

= 258.92 \m

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