a man walks on an equilateral triangle along path ABC with constant speed then the ratio of average speed and magnitude of average velocity A to C .
Answers
Answer:
the final answer is 2
Explanation:
since it's an equilateral triangle
let all the sides be=d
avg speed= d+d/t
2d/t (1)
avg velocity = d/t (2)
taking ratio of (1) and (2)
we get,, the answer 2
Given: Equilateral triangle ΔABC (assume side of Δ = a, all sides are equal)
speed of man = constant
To Find: Average speed - S/[Average velocity ]-v (A to C)
Solution:
Formula used:
v = Displacement(s) / time (t), Displacement is the shortest distance travelled.
S= Total distance/t
Applying the above formulas:
Total distance (A→C) = a + a = 2a ( according to the given image)
S = 2a/t
Total displacement (A→C) - s = a( single side AC of Δ)
[v] = a/t (only magnitude)
S/[v] = 2a × t/a × t
= 2a/a
= 2
Hence, the ratio of average speed and magnitude of average velocity A to C is 2