Car A is 50km east of car B and begins moving west at 50km/h. Simultaneously, car B begins moving north at 90km/h. Find:

Question

a) The closest distance the cars will be from each other in kilometers b) What time "t" in hours will this distance occur

So this is a practice question from my grade 12 calculus textbook. Usually I'm pretty competent at DVA problems, however, with this one I'm having trouble mainly with how to approach it. I've done problems where I have to find the rate of change of the distance between two objects as well as DVA problems on finding maximum and minimum values, but this one is confusing me a little as to how to approach it. I'm guessing maybe I need to start by finding an expression for each vehicles speed, than use that to maybe find a point of intersection maybe? Or maybe I should start with the general position formula or maybe the Pythagorean theorem. But I really don't know and could use a little help to get going here please and thanks.

Answer 1

hint

In the repere $ (B, west-east, south-north) $, At the instant $ t $, the car $ B $ will have coordinates $$B(0,V_b.t)$$ while the car $ A $, is at the point $$A(50-V_a.t,0)$$

the square of the distance between the two cars is $$AB^2=(50-V_a.t)^2+(V_b.t)^2$$ $$=2500-100V_a.t+(V_a^2+V_b^2)t^2$$

the minimum of this distance is attained when the derivative is zero: $$2t(V_a^2+V_b^2)-100V_a=0$$

thus

$$t=\frac{100V_a}{2(V_a^2+V_b^2)}$$ $$=\frac{5000}{2(2500+8100)}=$$

Answer 2

Pretend the cars are on a 2-D plane and arbitrarily place car B at the origin, i.e. (0,0). Car A is east of car B, so that puts car A on the x-axis at (50, 0). This is good for visualization.

Now, according to the problem, car B is moving north, i.e. up along the y-axis at 90 km/h and car A is moving west, i.e. left along the x-axis at 50 km/h.

At some time $t$, car B will be at $(0, 90t)$ and car A will be at $(50-50t, 0)$. This is simply using distance = speed X time. Given the way car B is moving, it's x co-ordinate always stays 0. Similarly, the y co-ordinate of car A always stays 0.

Now let's compute the distance between them at time $t$. The squared distance between two points $(a, b)$ and $(c, d)$ is given by $(a-c)^2 + (b-d)^2$. By this token, the squared distance between car A and car B is:

$d(t) = (50-50t)^2 + (90t)^2$.

To minimize this distance, let's take derivative with respect to $t$ and set to 0.

$d'(t) = 2(50-50t)(-50) + 2 \times 90t \times 90 = 0$.

You can solve to $t = 25/106$ hours. Finally, you can plug in $25/106$ into $d(t)$, take square root, and get the minimum distance.

Note that it's easier to work with squared distance when taking derivative. Any $t$ that minimizes the distance will also minimize the squared distance, so this is OK.