STEM Problem Detail

Problem #28

Mechanical Engineering Hard

Problem Statement

Prompt

Provide step by step solution of question 11.

Solution Rendering...

## Problem 21

Submarine speed = 10 mph SW to NE: velocity \( \vec{v}_s = 10(\cos 45^\circ, \sin 45^\circ) = \frac{10}{\sqrt{2}}(1,1) \) mph.

Steamer course due W to E: velocity \( \vec{v}_m = (V, 0) \) with \( V \) unknown.

Relative bearing: steamer always NW of submarine. Let submarine position \( \vec{r}_s \), steamer position \( \vec{r}_m \). Relative position \( \vec{r}_m - \vec{r}_s \) is always NW, i.e., direction vector \( (-1,1) \). So \( \vec{r}_m - \vec{r}_s = k(-1,1) \) for some \( k>0 \).

Differentiate: \( \vec{v}_m - \vec{v}_s = \frac{dk}{dt}(-1,1) \).

So \( (V,0) - \frac{10}{\sqrt{2}}(1,1) = \lambda(-1,1) \) for some scalar \( \lambda \).

Equate components:
\( V - \frac{10}{\sqrt{2}} = -\lambda \),
\( -\frac{10}{\sqrt{2}} = \lambda \).

Thus \( \lambda = -\frac{10}{\sqrt{2}} \), then \( V = \frac{10}{\sqrt{2}} - \frac{10}{\sqrt{2}} = 0 \). That would mean steamer stationary? Contradiction.

Re-examine: "steamer bears N.W. of the submarine all the while" means the line from submarine to steamer is always in the NW direction. So relative position vector is proportional to NW direction. But that doesn't mean relative velocity is along NW; it means the relative position maintains direction, so the relative velocity must be along that same line (otherwise direction changes). So \( \vec{v}_m - \vec{v}_s \) is along NW.

Thus \( (V,0) - \frac{10}{\sqrt{2}}(1,1) = \mu(-1,1) \).

Then \( V - \frac{10}{\sqrt{2}} = -\mu \) and \( -\frac{10}{\sqrt{2}} = \mu \).

So \( \mu = -\frac{10}{\sqrt{2}} \), \( V = \frac{10}{\sqrt{2}} - \frac{10}{\sqrt{2}} = 0 \) again. This suggests the steamer is stationary relative to ground, which seems odd. Possibly the interpretation: "bears N.W." means the bearing is constant at NW, so the relative velocity is zero along the perpendicular direction? Actually, if bearing is constant, the relative velocity vector must be along the line of bearing. So set up: Let submarine at origin initially, steamer at \( (x_0, y_0) \) with \( y_0 > 0, x_0 < 0 \) for NW. Then \( \vec{r}_m(t) = (x_0 + V t, y_0) \), \( \vec{r}_s(t) = \frac{10}{\sqrt{2}} t (1,1) \). Relative position \( \vec{r}_m - \vec{r}_s = (x_0 + V t - \frac{10}{\sqrt{2}} t, \; y_0 - \frac{10}{\sqrt{2}} t) \).

For bearing always NW, the ratio of components must be constant: \( \frac{y_0 - \frac{10}{\sqrt{2}} t}{x_0 + V t - \frac{10}{\sqrt{2}} t} = \frac{1}{-1} = -1 \) (since NW direction vector (-1,1) gives slope -1). So \( y_0 - \frac{10}{\sqrt{2}} t = -x_0 - V t + \frac{10}{\sqrt{2}} t \).

Simplify: \( y_0 + x_0 = \frac{20}{\sqrt{2}} t - V t \). For this to hold for all \( t \), coefficients must match: LHS constant, so \( \frac{20}{\sqrt{2}} - V = 0 \) ⇒ \( V = \frac{20}{\sqrt{2}} = 10\sqrt{2} \approx 14.14 \) mph, and \( y_0 + x_0 = 0 \).

Thus steamer speed east = \( 10\sqrt{2} \) mph.

Now, submarine stops when distance to steamer = 7 miles. Let that time be \( t_1 \). At \( t_1 \), relative position \( \vec{r}_m - \vec{r}_s = (x_0 + V t_1 - \frac{10}{\sqrt{2}} t_1, \; y_0 - \frac{10}{\sqrt{2}} t_1) \). Using \( y_0 = -x_0 \), and \( V = \frac{20}{\sqrt{2}} \).

Compute: \( x_0 + \frac{20}{\sqrt{2}} t_1 - \frac{10}{\sqrt{2}} t_1 = x_0 + \frac{10}{\sqrt{2}} t_1 \), and \( -x_0 - \frac{10}{\sqrt{2}} t_1 = -(x_0 + \frac{10}{\sqrt{2}} t_1) \). So relative position vector is \( (X, -X) \) with \( X = x_0 + \frac{10}{\sqrt{2}} t_1 \). Distance = \( \sqrt{X^2 + (-X)^2} = \sqrt{2} |X| = 7 \) ⇒ \( |X| = \frac{7}{\sqrt{2}} \).

After submarine stops, steamer continues east at speed \( V \). Submarine fires torpedo due north at 30 mph after delay \( \tau \). Let time of firing be \( t_1 + \tau \). At that instant, steamer position: \( \vec{r}_m(t_1+\tau) = (x_0 + V(t_1+\tau), \; y_0) \), submarine at \( \vec{r}_s(t_1) = \frac{10}{\sqrt{2}} t_1 (1,1) \) (since stopped). Torpedo from submarine position, velocity (0,30).

Let hit time after firing = \( T \). Then torpedo position at hit: \( \vec{r}_T = \vec{r}_s(t_1) + (0,30) T \). Steamer position: \( \vec{r}_m(t_1+\tau+T) = (x_0 + V(t_1+\tau+T), \; y_0) \). For hit, positions equal: \( x_0 + V(t_1+\tau+T) = \frac{10}{\sqrt{2}} t_1 \) (x-coordinate), and \( y_0 + 30T = \frac{10}{\sqrt{2}} t_1 \) (y-coordinate).

From earlier, \( y_0 = -x_0 \), \( V = 10\sqrt{2} \), and \( \frac{10}{\sqrt{2}} t_1 + x_0 = X = \pm \frac{7}{\sqrt{2}} \).

Use x-equation: \( x_0 + 10\sqrt{2}(t_1+\tau+T) = \frac{10}{\sqrt{2}} t_1 \). Rearrange: \( x_0 + 10\sqrt{2} t_1 + 10\sqrt{2}(\tau+T) = \frac{10}{\sqrt{2}} t_1 \). But \( \frac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07 \). Wait: \( 10\sqrt{2} \approx 14.14 \), \( \frac{10}{\sqrt{2}} \approx 7.07 \). So: \( x_0 + 14.14 t_1 + 14.14(\tau+T) = 7.07 t_1 \). ⇒ \( x_0 + 14.14 t_1 - 7.07 t_1 + 14.14(\tau+T) = 0 \) ⇒ \( x_0 + 7.07 t_1 + 14.14(\tau+T)=0 \). (1)

y-equation: \( -x_0 + 30T = 7.07 t_1 \). (2)

Add (1) and (2): \( (x_0 - x_0) + 7.07 t_1 + 14.14(\tau+T) + 30T = 7.07 t_1 \) ⇒ \( 14.14\tau + 14.14T + 30T = 0 \) ⇒ \( 14.14\tau + 44.14T = 0 \) ⇒ impossible since all positive. Maybe sign of X: We had \( X = x_0 + \frac{10}{\sqrt{2}} t_1 = \pm \frac{7}{\sqrt{2}} \). Try negative: \( X = -\frac{7}{\sqrt{2}} \). Then \( x_0 = -\frac{7}{\sqrt{2}} - \frac{10}{\sqrt{2}} t_1 = -\frac{1}{\sqrt{2}}(7+10 t_1) \).

Now x-equation: \( x_0 + 10\sqrt{2}(t_1+\tau+T) = \frac{10}{\sqrt{2}} t_1 \).

Substitute \( x_0 \): \( -\frac{1}{\sqrt{2}}(7+10 t_1) + 10\sqrt{2}(t_1+\tau+T) = \frac{10}{\sqrt{2}} t_1 \).

Multiply by \( \sqrt{2} \): \( -(7+10 t_1) + 20 (t_1+\tau+T) = 10 t_1 \).

Simplify: \( -7 -10 t_1 + 20 t_1 + 20\tau + 20T = 10 t_1 \) ⇒ \( -7 + 10 t_1 + 20\tau + 20T = 10 t_1 \) ⇒ \( -7 + 20\tau + 20T = 0 \) ⇒ \( 20(\tau+T) = 7 \) ⇒ \( \tau+T = 0.35 \) hours = 21 minutes.

y-equation: \( y_0 + 30T = \frac{10}{\sqrt{2}} t_1 \). But \( y_0 = -x_0 = \frac{1}{\sqrt{2}}(7+10 t_1) \). So \( \frac{1}{\sqrt{2}}(7+10 t_1) + 30T = \frac{10}{\sqrt{2}} t_1 \).

Multiply by \( \sqrt{2} \): \( 7+10 t_1 + 30\sqrt{2} T = 10 t_1 \) ⇒ \( 7 + 30\sqrt{2} T = 0 \) ⇒ \( T = -\frac{7}{30\sqrt{2}} \) negative. So impossible.

Thus perhaps the interpretation is different. Maybe "bears N.W." means the line from submarine to steamer is in NW direction, but not necessarily constant? Alternatively, perhaps the submarine's course is SW to NE, meaning its velocity is in that direction, but the bearing condition might hold initially only? The problem likely intends that the relative velocity is such that the bearing remains constant. Typically, for constant bearing, the relative velocity must be along the line joining them. So set up: Let initial positions: submarine at (0,0), steamer at (-d, d) for some d>0 (NW direction). Let submarine velocity \( \vec{v}_s = (10/\sqrt{2}, 10/\sqrt{2}) \), steamer velocity \( (V,0) \). Then relative position at time t: \( \vec{r}_m - \vec{r}_s = (-d + V t - \frac{10}{\sqrt{2}} t, \; d - \frac{10}{\sqrt{2}} t) \). For bearing NW, the ratio of components must equal 1/-1? Actually NW vector has negative x, positive y, so ratio y/x = -1. So \( \frac{d - \frac{10}{\sqrt{2}} t}{-d + V t - \frac{10}{\sqrt{2}} t} = -1 \) ⇒ \( d - \frac{10}{\sqrt{2}} t = d - V t + \frac{10}{\sqrt{2}} t \) ⇒ \( - \frac{10}{\sqrt{2}} t = - V t + \frac{10}{\sqrt{2}} t \) ⇒ \( V t = \frac{20}{\sqrt{2}} t \) ⇒ \( V = \frac{20}{\sqrt{2}} = 10\sqrt{2} \), independent of d. So same as before.

Then at time t1 when distance=7, we have \( \vec{r}_m - \vec{r}_s = (-d + V t_1 - \frac{10}{\sqrt{2}} t_1, \; d - \frac{10}{\sqrt{2}} t_1) = ( -d + \frac{10}{\sqrt{2}} t_1, \; d - \frac{10}{\sqrt{2}} t_1 ) \). Distance squared: \( (-d + \frac{10}{\sqrt{2}} t_1)^2 + (d - \frac{10}{\sqrt{2}} t_1)^2 = 2(d - \frac{10}{\sqrt{2}} t_1)^2 = 49 \) ⇒ \( |d - \frac{10}{\sqrt{2}} t_1| = \frac{7}{\sqrt{2}} \). Choose sign so that meeting possible later? Assume d - \frac{10}{\sqrt{2}} t_1 = \frac{7}{\sqrt{2}} (positive). Then \( d = \frac{7}{\sqrt{2}} + \frac{10}{\sqrt{2}} t_1 \).

Now after submarine stops, positions at firing time t1+τ: submarine at \( \vec{r}_s(t_1) = (\frac{10}{\sqrt{2}} t_1, \frac{10}{\sqrt{2}} t_1) \), steamer at \( \vec{r}_m(t_1+\tau) = (-d + V(t_1+\tau), d) = ( -\frac{7}{\sqrt{2}} - \frac{10}{\sqrt{2}} t_1 + 10\sqrt{2}(t_1+\tau), \; \frac{7}{\sqrt{2}} + \frac{10}{\sqrt{2}} t_1 ) \).

Torpedo fired due north from submarine position at speed 30. Let time to hit after firing = T. Torpedo position at hit: \( ( \frac{10}{\sqrt{2}} t_1, \; \frac{10}{\sqrt{2}} t_1 + 30T ) \). Steamer at hit: \( ( -\frac{7}{\sqrt{2}} - \frac{10}{\sqrt{2}} t_1 + 10\sqrt{2}(t_1+\tau+T), \; \frac{7}{\sqrt{2}} + \frac{10}{\sqrt{2}} t_1 ) \).

Equate x-coordinates: \( \frac{10}{\sqrt{2}} t_1 = -\frac{7}{\sqrt{2}} - \frac{10}{\sqrt{2}} t_1 + 10\sqrt{2}(t_1+\tau+T) \).

Multiply by \( \sqrt{2} \): \( 10 t_1 = -7 - 10 t_1 + 20(t_1+\tau+T) \) ⇒ \( 10 t_1 = -7 -10 t_1 + 20 t_1 + 20\tau + 20T \) ⇒ \( 10 t_1 = -7 + 10 t_1 + 20\tau + 20T \) ⇒ \( 0 = -7 + 20\tau + 20T \) ⇒ \( \tau+T = \frac{7}{20} = 0.35 \) hours = 21 minutes.

Equate y-coordinates: \( \frac{10}{\sqrt{2}} t_1 + 30T = \frac{7}{\sqrt{2}} + \frac{10}{\sqrt{2}} t_1 \) ⇒ \( 30T = \frac{7}{\sqrt{2}} \) ⇒ \( T = \frac{7}{30\sqrt{2}} \approx 0.165 \) hours ≈ 9.9 minutes.

Then \( \tau = 0.35 - 0.165 = 0.185 \) hours ≈ 11.1 minutes.

So submarine must fire about 11.1 minutes after stopping.

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