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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?

User Will Ryan
by
5.5k points

1 Answer

3 votes

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
u=-7 m/s

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

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

we will use the equation of motion:


v=u+at

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:


\Rightarrow v=-7m/s+6 m/s^2 * 4 s=24m/s-7 m/s=17 m/s

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


User Iwazovsky
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6.3k points