Question

ABCD is a square of side 100m. A cyclist P starts from A and moves towards B with a speed of 4 m/s. At the same instant, a second cyclist Q starts from B and moves towards C with speed of 3 m/s. If the angular velocity of Q relative to P after t seconds is $\frac{4sin\theta +ncos\theta }{\sqrt{(100-4t{)}^2+n{t}^2}}$ rad/s, find value of n.

Answer 1

At any instant of time t after the start
$AP=4t$
$BQ=3t$
$PB=100-4t$
Velocity of P perpendicular to PQ is 4 sin $\theta$
Velocity of Q perpendicular to PQ is 3 cos $\theta$
Relative velocity of Q with respect to P is
$4sin\theta -(-3cos\theta )=4sin\theta +3cos\theta$
The angular velocity of Q with respect to P is
=$\frac{4sin\theta +3cos\theta }{PQ}$
$=\frac{4sin\theta +3cos\theta }{\sqrt{P{B}^2+B{Q}^2}}$
$=\frac{4sin\theta +3cos\theta }{\sqrt{(100-4t{)}^2+9t^2}}$ rad/s