Question

The average speed of vehicles along a circular track is 36kmph , whose radius of a curvature is 20m. The banking angle required for the curve is
A. tan-1 (2)
B. tan-1 (1/2)
C. tan-1 (2/3)
D. tan-1 (3/2)​

Answer 1

Answer:

B)tan-1(1/2)

Explanation:

As in the case of banking of road,

tanx=v2/rg...............(v=speed of car,r=radius of curvature,g=10m/s2)

here,

v=36km/hr=10m/s

v=36km/hr=10m/sr=20m

v=36km/hr=10m/sr=20mg=10m/s2

.....

therefore,

tanx=(10)^2/20×10

tanx=1/2

i.e., x=tan-1(1/2)

Answer 2

Gɪᴠᴇɴ

$\tt \leadsto Time(t_{1}) = 1st \: sec \\\\ \tt \leadsto{}Time(t_{2}) = 3rd \: sec\\\\ \tt \leadsto{}Distance(s_{1}) = 3 \: m \\\\ \tt \leadsto{} Distance(s_{2}) = 7 \: m$

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ᴛᴏ ꜰɪɴᴅ

➤ The banking angle required for the curve?

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Sᴛᴇᴘꜱ

$\tt \mapsto s_{1}= u + \frac{a}{2} (2 n_{1} - 1) \\ \\ \tt\mapsto 3 = u + \frac{a}{2} (2 \times 1 - 1) \\ \\ \tt \mapsto 3 = u + \frac{a}{2} \\ \\ \tt \mapsto 2u + a = 6 - - - - - (1) \\ \\ \\ \\ \\ \tt\mapsto s_{2}= u + \frac{a}{2} (2 n_{2} - 1) \\ \\ \tt \mapsto 7 = u + \frac{a}{2} (2 \times 3 - 1) \\ \\ \tt \mapsto7 = u + \frac{a}{2} \times 5 \\ \\ \tt \mapsto 14 = 2u + 5a \\ \\ \tt \mapsto{}2u + 5a= 14 - - - - - (2) \\ \\ \text{Subtracting \: (1) \: from \: (2)} \\ \tt \leadsto 5a - a = 14 - 6 \\ \\ \tt \leadsto 4a = 8 \\ \\ \green{\tt \leadsto{}a =2 \: m/{s}^{2} } \\ \\ \text{Putting \: value \: of \: a \: in \: (1)} \\ \tt\implies 2u + 2 = 6 \\ \\ \tt \implies 2u = 4 \\ \\ \green{\tt \implies u = 2 \: m/s}$

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