# A motor car is going due north at a speed of 50 km/h. It makes a 90 degree left turn without changing the speed . the change in velocity of the car is about?

## 50 km/h towards west 70 km/h towards south-west 70 km/h towards north-west zero

Jun 29, 2018

70 km/h towards south-west

#### Explanation:

I suggest thinking of this being a question about an airplane. Let's rewrite the question:

An airplane is going due north at a speed of 50 km/h. A strong wind comes up suddenly which makes the plane's progress, as seen from the ground, make a 90 degree left turn without changing the speed. the change in velocity of the plane is about?

Clearly this is a question about vector addition. The plane's velocity, with respect to the air it is flying in, is 50 km/h to the North. But the plane is observed from the ground to be pointed north but going west at 50 km/h. It must be that a wind is blowing and carrying the plane off course. What must the direction and speed of the wind be to explain this?

OK we have 2 component velocity vectors and 1 resultant vector to analyse. The plane's speed thru the air is one component: 50 km/h to the North. The other component is the wind velocity -- our unknown. The resultant is 50 km/h to the West.

Make a drawing as follows (I will explain after you get the drawing made). Draw the 50 km/h to the North vector. Starting from the tail of that North vector, draw the 50 km/h to the West vector. Now draw an arrow starting from the arrowhead of the 50 km/h to the North vector to the arrowhead of the west vector. You have a ${45}^{\circ}$ right triangle! The hypotenuse is our wind velocity.

You may see that this looks like one of the graphical methods to draw a resultant vector. You started with the one component and the resultant. But that does not matter. Now that it is drawn, we can use trigonometry to find out what the wind speed is. Look at the angle where the 2 arrowheads are, we can write the sine equation of that angle.

$\sin {45}^{\circ} = \text{opposite"/"hypotenuse}$

$\sin {45}^{\circ} = 0.707 = \frac{50 k \frac{m}{h}}{\text{hypotenuse}}$

$\text{hypotenuse} = \frac{50 k \frac{m}{h}}{0.707} = 70.7 k \frac{m}{h}$

You can see from your drawing that the direction of the wind is South-West.

Thinking about the original question, the "change in velocity" of the car would be the velocity that must be added to the original velocity to yield the new velocity.

I hope this helps,
Steve