Question

An object covers (16.0 ± 0.4)m distance in (4.0 ± 0.2). Find out its speed.

Answer 1

"The object covers (16.0 ± 0.4) m distance with (4.0 ± 0.2).

V = 13.8/4 = 3.45

Δvv = Δss + Δtt = 0.213.8 + 0.34

⇒Δv= (0.213.8+0.34) ×3.45= 0.3

Solve this using an equation of motion called v = u + at

U = 3i+4j

a = 0.3j+0.4i

t = 10

v = 4j+3i+10(0.3j +0.4i)

v = 7i+7j

|v|= {2(49)} ^1/2

|v|= 7(2^1/2)

"

Answer 2

$7 \sqrt{2}$

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