A tiger chases a deer 30m ahead of it and gains 3m in first 5s after chase started.After 10s,the distance b/n them is (both initially at rest)
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hii mate!!
=> relative initial velocity(assuming both start from rest for a neutral ground observer) = vₐ - vᵦ = 0
relative acceleration = aₐ - aᵦ
relative distance covered in 5 s (deer at relative rest) = 3 m
=> s = ut + ½at²
=> 3 = 0t + ½[aₐ - aᵦ](5)²
=> aₐ - aᵦ = relative acceleration = 3/12.5 = 0.24 m/s²
=> the distance gained by the tiger in 10sec(neutral observer) = 0t + ½[aₐ - aᵦ](10)² = .24 x 50 = 12 m
hope this help☺☺☺☺
=> relative initial velocity(assuming both start from rest for a neutral ground observer) = vₐ - vᵦ = 0
relative acceleration = aₐ - aᵦ
relative distance covered in 5 s (deer at relative rest) = 3 m
=> s = ut + ½at²
=> 3 = 0t + ½[aₐ - aᵦ](5)²
=> aₐ - aᵦ = relative acceleration = 3/12.5 = 0.24 m/s²
=> the distance gained by the tiger in 10sec(neutral observer) = 0t + ½[aₐ - aᵦ](10)² = .24 x 50 = 12 m
hope this help☺☺☺☺
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