What is the average velocity of the deer during startlement?

Question

A startled deer runs 520 m at 20 degrees north of east for half a minute, then turns and runs 380 m at 55 degrees north of west for 15 seconds and stops. what is the average velocity of the deer during this time?

Answer 1

For this solution I am going to round to the nearest whole number.

The key to solving this problem comes from evaluating the total distance traveled in the north, east and west. This is done by using right triangle trig. COS(x) = adjacent/hypotenuse, SIN(x) = opposite/hypotenuse

For this problem since we are given the hypotenuse, we can rearrange the above equations as follows (multiply both sides by the hypotenuse)

hypotenuse*COS(x) = adjacent

hypotenuse*SIN(x) = opposite

For the first part of our motion we can set up a right triangle with the hypotenuse equal to 520m at 20 degrees above horizontal.

So we can calculate the eastern motion as:

520*COS(20) = 489

And the northern motion

520*SIN(20) = 178

For the second part we set up a right triangle with the hypotenuse equal to 380m at 55 degrees above horizontal.

So we calculate the western motion as:

380*COS(55) = 218

and the northern motion

380*SIN(55) = 311

Then we can combine the total vertical distance traveled at 178 + 311 = 489

Then since the horizontal motion is in opposite directions we must find the difference:

489 - 218 = 271

We can now set up a right triangle with the horizontal leg equal to 271 and the vertical leg equal to 489.

When we use tangent we can find the angle

TAN(x) = opposite/adjacent

TAN(x) = 489/271

x = TAN^-1(489/271) = 61 -> 61 degrees north of east

If we use SIN(x), COS(x) or the pythagorean theorem we can find the total displacement. Here we will use SIN(x)

SIN(61) = 489/hypotenuse

hypotenuse = 489/(SIN(61)) = 559 meters

When we divide 559 meters by the total time of 45 seconds that gives us 12 m/s.

Putting this all together we get 12 m/s at 61 degrees north of east.