Acanoe is drifting left toward a hungry hippo with a velocity of 7\,\dfrac{\text{m}}{\text{s}}7 s m 7, space, start fraction, m, divided by, s, end fraction. the canoe rider starts paddling frantically, causing the canoe to travel to the right with a constant acceleration of 6 \,\dfrac{\text m}{{\text s}^2}6 s 2 m 6, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. after 4\,\text s4s4, space, s, what is the velocity of the canoe?

17 m/s

Velocity and acceleration are vector quantities. Let leftwards be defines as negative and rightwards as positive direction.

It is given that canoe is drifting left with a velocity of

The canoe rider is moving towards right with a constant acceleration,

we need to find the velocity of canoe after 4 s.

we will use the equation of motion:

where, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Insert the values in the above equation:

Hence, velocity of canoe after 4 s would be 17 m/s.

massxgravity x heght=weightxheight= pe

4900=490xh

h=10

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min