Question

Please please please please please helpp ...

Answer 1

$\frac{dr1}{dt} = v1 = 3i + 8t \: j \\ \frac{dr2}{dt} =v2 = 8t \: i + 3j \\ at \: t = 1 \\v1 = 3i + 8j \\ and \\ v2 = 8i + 3j \\ therefore \\ \\ the \: relative \: velocity \: is \\ v1 - v2 = - 5i + 5j \\ magnitude = \sqrt{( - 5) {}^{2} + 5 {}^{2} } = 5 \sqrt{2} \: m.s {}^{ - 1}$

Answer 2

Explanation:

v1 at t=1 second will be 3î+8j

v2 at t= 1 second will be 8î+3j

V relative = V2- V1

=5î-5j

magnitude of V relative=5√2 m/s