A canoe 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?

Question

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Respuesta :

ariston ariston · Answer 1 · 2018-10-03T08:54:51+02:00

Answer: 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 [tex]u=-7 m/s[/tex]

The canoe rider is moving towards right with a constant acceleration, [tex]a=6 m/s^2[/tex]

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

we will use the equation of motion:

[tex]v=u+at[/tex]

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:

[tex]\Rightarrow v=-7m/s+6 m/s^2 \times 4 s=24m/s-7 m/s=17 m/s[/tex]

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