A70 m wide river flows at 0.80 m/s [e]. a girl, starting on the south shore, swims across it at 1.4 m/s relative to the water.

a) what is the least time she requires to cross the river?

b) how far downstream will she be when she lands on the opposite shore?

c) at what angle to the shore would she have to aim, in order to arrive at a point directly opposite the starting point?

50 seconds

t = d /v

where t is the time taken, d is the distance and v is the speed

d= 70 m and s= 1.4 m/s

Therefore;

t = 70/1.4

 = 50 seconds

40 meters

dc = Vc × t

Where dc is the distance down stream, Vc is the velocity downstream and t is the time taken

Vc = 0.8 m/s

t = 50 seconds

Therefore;

dc = 0.8 × 50

    = 40 meters

55 degrees

cosA = 0.8/1.4  

A = 55.15 ° with the horizontal (shore) upstream  

Therefore; the angle to aim from the shore is 55°

f=ma newton's second law 7x8=56n