A man of mass 60 kg is standing on a boat of mass 140 kg, which is at rest in still water. The man is initially at 20 m from the shore. He starts walking on the boat for 4 s with constant speed 1.5 m / s towards the shore. The final distance of the man from the shore is

Question

A man of mass 60 kg is standing on a boat of mass 140 kg, which is at rest in still water. The man is initially at 20 m from the shore. He starts walking on the boat for 4 s with constant speed 1.5 m / s towards the shore. The final distance of the man from the shore is (A) 15.8 m (B) 4.2 m (C) 12.6 m (D) 14.1 m

Tardigrade Admin · Accepted Answer

Distance travelled by the man on boat in 4 second =(1.5) × 4 = 6.0 m 140 x=60(6-x) 140 x=360-60 x x=1.8 m So final distance of the man from the shore will be 20-(6-1.8)=15.8 m

Q. A man of mass $60 k g$ is standing on a boat of mass $140 k g$ , which is at rest in still water. The man is initially at $20 m$ from the shore. He starts walking on the boat for $4 s$ with constant speed $1.5 m / s$ towards the shore. The final distance of the man from the shore is

15.8 m

4.2 m

12.6 m

14.1 m

Distance travelled by the man on boat in 4 second $= (1.5) \times 4 = 6.0 m$
$140 x = 60 (6 - x)$
$140 x = 360 - 60 x$
$x = 1.8 m$
So final distance of the man from the shore will be
$20 - (6 - 1.8) = 15.8 m$

Q. A man of mass 60kg is standing on a boat of mass 140kg, which is at rest in still water. The man is initially at 20m from the shore. He starts walking on the boat for 4s with constant speed 1.5m/s towards the shore. The final distance of the man from the shore is