As these are distances created by moving in a straight line, using a trigonometric analysis can solve the missing single straight-line displacement. Looking at the 48m and 12m movements as legs of a triangle, obtaining the hypotenuse using the pythagorean theorem will yield us the correct answer.
This is shown below:
c^2 = 48^2 + 12^2
c = sqrt(2304 + 144)
c = sqrt(2448)
c = 49.48 m
To obtain the angle at which Anthony walks 49.48, we obtain the arc tangent of (12/48). This is shown below:
arc tan (12/48) =14.04 degrees.
Therefore, Anthony could have walked 49.48 m towards the S 14.04 W direction.
Answer:
48.4 km, 34.3° north of east
Explanation:
Let's say east is the +x direction and north is the +y direction.
Adding up the x components of the vectors:
x = 20 cos 60 + 30 + 0
x = 40 km
Adding up the y components of the vectors:
y = 20 sin 60 + 0 + 10
y = 27.3 km
The magnitude of the displacement is:
d = √(x² + y²)
d = 48.4 km
The direction is:
θ = atan(y/x)
θ = 34.3° north of east
Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above
Answer:
using a formula
average velocity=total displacement/total time
average velocity=120/1minute40sec
1minute=60sec
60sec+40sec=120sec
average velocity=120m/120sec
average velocity=1m/sec
