Answer:
a) v1 = 5.52m/s
b) v2 = -1.52m/s
c) v3 = 4.62m/s
d) vt = 3.85m/s
Explanation:
The velocity of the football wide receiver is his displacement per unit time.
Velocity v = (displacement d)/time t
v = d/t .....1
For each of the cases, equation 1 would be used to calculate the velocity.
a) v1 = d1/t1
d1= 16m
t1 = 2.9s
v1 = 16m/2.9s
v1 = 5.52m/s
b) v2 = d2/t2
d2 = -2.5m
t2 = 1.65s
v2 = -2.5/1.65
v2 = -1.52m/s
c) v3 = d3/t3
d3 = 24m
t3 = 5.2s
v3 = 24/5.2
v3 = 4.62m/s
d) vt = dt/tt
dt = 16m - 2.5m + 24m = 37.5m
tt = 2.9 + 1.65 + 5.2 = 9.75s
vt = 37.5/9.75
vt = 3.85m/s
Seven
The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer
Eight
They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.
The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.
I really do object to the wording, but what can I do?
Nine
Nine is the same thing as 8.
Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408
The actual distance is 400 + 80 = 480
The difference is the answer = 480 - 408 = 72 <<<< Answer
Ten
This is just the displacement magnitude.
dis = sqrt(30^2 + 80^2)
dis = sqrt(900 + 6400)
dis = sqrt(7300)
dis = 85.44 <<<< Answer D
Twelve
Vi = 2.15*Sin(30) = 1.075 m/s
vf = 0
a = - 9.81
t = ?
<u>Formula</u>
a = (vf - vi)/t
<u>Solve</u>
-9.81 = (0 - 1.075)/t
- 9.81 * t = -1.075
t = 0.11 seconds
Thirteen
I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.
Edit
That is a surprise! Really quickly
d = 3.2 m
a = - 9.82
vf = 0
vi = ?
vf^2 = vi^2 - 2*a*d
0 = vi^2 - 2*9.81*3.2
vi = sqrt(19.62*3.2)
vi = 8.0 m/s But that is the vertical component of the speed
v = vi/sin(25)
v = 8.0/sin(25) = 11
30 grams of radioactive isotope have passed.
