A tiger chases a deer 30m ahead of it and gains 3m in 5s. After the chase starts. After 10s the distance between them is
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[please note: if the the velocities of objects A(cat) and B(deer) are vₐ and vᵦ with respect to a stationary observer (neutral observer) on the ground surface, then the velocity of A relative to B is vₐ - vᵦ. this is as if B were at rest and only A were moving.]
=> 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 helps
=> 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 helps
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