Question

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorboat? b) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? c) What distance downstream does the boat reach the opposite shore?

Answer 1

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

Answer 2

Answer:

The distance downstream does the boat the opposite shore is $\mathrm{x}=\mathbf{4 8} meters$

Explanation:

The velocity given in the east is $v_{1}=4 \mathrm{~m} / \mathrm{s}$

The velocity given in the north is $v_{2}=3 \mathrm{~m} / \mathrm{s}$

a). The resultant velocity :

$v=\sqrt{4^{2}+3^{2}}$

$=5 \mathrm{~m} / \mathrm{s}$

The river is 80-meters wide. That is, the distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation.

(b) The width of the river:

  $d=80m$

$t=\frac{d}{v}$

$t=\frac{80 m}{5 m / s}$

$t=16 seconds.$

(c) Distance downstream does the boat reach the opposite shore:

$x=v_{2} \times t$

$x=3 \mathrm{~m} / \mathrm{s} \times 16 \mathrm{~s}$

$x=48 meters.$

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