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Running at maximum speed, it takes a boat times as long to go 10 miles upstream as it does to go 10 miles downstream. If the current is 4 mph, find the speed of the boat in still water.

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Note: The question states the time to go upstream is a number of times (not explicitly written) the time to go downstream. We'll assume a general number N

Answer:


\displaystyle v_b=(N+1)/(N-1)(4\ mph)

Step-by-step explanation:

Relative Speed

If a boat is going upstream against the water current, the true speed of motion is
v_b-v_w, being
v_b the speed of the boat and
v_w the speed of the water. If the boat is going downstream, the true speed becomes
v_b+v_w.

The question states the time to go upstream is a number of times N (not explicitly written) the time to go downstream. The speed of an object is computed as


\displaystyle v=(x)/(t)

Where x is the distance traveled and t the time taken for that. The time can be computed by


\displaystyle t=(x)/(v)

If
t_u is the time for the upstream travel and
t_d is the time for the downstream travel, then


t_u=Nt_d

Siince the same distance x= 10 miles is traveled in both cases:


\displaystyle (10)/(v_b-v_w)=N(10)/(v_b+v_w)

Simplifying and rearrangling


v_b+v_w=N(v_b-v_w)

Operating


v_b+v_w=Nv_b-Nv_w

Solving for
v_b


\displaystyle v_b=(N+1)/(N-1)v_w


If\ N=2,\ v_w=4\ mph


\displaystyle v_b=(3)/(1)(4)=12\ mph

If N=3


\displaystyle v_b=(4)/(2)v_w=2(4)=8\ mph

We can use the required value of N to compute the speed of the boat as explained

User Amrut Bidri
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