BMC Test 2 Math || || ||
Study ||online ||at ||https://quizlet.com/_d1wvh9
1. A canoeist accelerating uniformly from rest reaches a velocity of 3.5 ms-1
|| || || || || || || || || || ||
in just 5 seconds.
|| || || ||
a. what is the canoeist acceleration over this time period?
|| || || || || || || ||
b. assuming the acceleration is constant,what would be the acceleration after 10 | | | | | | | | | | ||
seconds?: a.
|| ||
(final velocity - initial velocity/change in time)
|| || || || || ||
3.5- = .7
|| || || || ||
.7 m/s^-2
||
b.
Final velocity = (initial velocity)+(acceleration)(time)
| || || ||
Vf = (0) + (.7) (10)
|| || || || || ||
=7 ms^-1 ||
2. A sprinter attempting a 60m dash completes the distance in a time of 6.5
|| || || || || || || || || || || || ||
seconds (starting fromrest).What is the sprinter's average velocity over this
|| || || | || || || || || || ||
distance?: Average velocity = change in speed / change in time
|| || || || || || || || || || ||
60-.5-0 = || || ||
9.23 m/s^-1 ||
3. A soccer ball is kicked from the ground with an initial velocity of 27 m/s, at
|| || || || || || || || || || || || || || ||
an angle of 39 degrees.
|| || || || ||
a. What is the horizontal and vertical component at release?: 27 cos(39 = 20.98
| | | | | | | | || | | |
|| horizontal = 20.98 m/s^-1 || || ||
27 sin(39 = 16.99
|| || ||
Vertical = 16.99 m/s^-1 || || ||
4. A ball was launched off a cliff with an initial velocity of 15 m/s at an angle
|| || || || || || || || || || || || || || || ||
of 315 degrees ccw from the horizontal (45 degrees below the horizontal).
|| || || || || || || || || || || ||
a. What is the horizontal and vertical component at release?: 15 cos(315 =
|| || || || || || || || || || ||
10.60
||
horizontal = 10.60 m/s^-1 15 || || || ||
|| sin(315 = -10.60 || ||
1 ||/
||8
, BMC Test 2 Math || || ||
Study ||online ||at ||https://quizlet.com/_d1wvh9
Vertical = -10.60 m/s^-1 || || ||
5. A baseball is thrown with a velocity of 15ms-1 at an angle of 30 degrees ccw
| | | | | | | | | | | | | | |
from the horizontal.
|| || ||
a. What are the horizontal and vertical components of the velocity just as the
|| || || || || || || || || || || ||
ball leaves thehand?
|| || ||
b. What is the horizontal and vertical velocity of the ball when it reaches
|| || || || || || || || || || || ||
maximum height?
|| ||
c. At what time does the ball reach maximum vertical height?
|| || || || || || || || ||
d. How far has the ball traveled vertically and horizontally at this time?: a.
|| || || || || || || || || || || ||
15 sin(30
|| ||
7.5 m/s initial vertical || || ||
15 cos(30 ||
12.99 m/s initial horizontal || || ||
b.
horizontal = 12.99 m/s || || ||
vertical = 0 m/s
|| || || ||
c.
final velocity = initial velocity + (acceleration)(time) 0
|| || || || || || ||
= 7.5 + (-9.81)(t)
|| || || ||
-7.5/-9.81 = .7645 seconds || || ||
|| d.
vertical displacement = (initial vertical velocity)(time)+(1/2)(-9.81)(time^2) x
|| || || || || ||
= (7.5) (.7645) + (1/2) (-9.81) (.7645^2)
|| || || || || || ||
= 2.87 M || ||
Horizontal displacement = (initial horizontal velocity)(time) x || || || || || ||
= (12.99) (.7645)
|| || ||
= 9.94 M || ||
2 ||/
||8
Study ||online ||at ||https://quizlet.com/_d1wvh9
1. A canoeist accelerating uniformly from rest reaches a velocity of 3.5 ms-1
|| || || || || || || || || || ||
in just 5 seconds.
|| || || ||
a. what is the canoeist acceleration over this time period?
|| || || || || || || ||
b. assuming the acceleration is constant,what would be the acceleration after 10 | | | | | | | | | | ||
seconds?: a.
|| ||
(final velocity - initial velocity/change in time)
|| || || || || ||
3.5- = .7
|| || || || ||
.7 m/s^-2
||
b.
Final velocity = (initial velocity)+(acceleration)(time)
| || || ||
Vf = (0) + (.7) (10)
|| || || || || ||
=7 ms^-1 ||
2. A sprinter attempting a 60m dash completes the distance in a time of 6.5
|| || || || || || || || || || || || ||
seconds (starting fromrest).What is the sprinter's average velocity over this
|| || || | || || || || || || ||
distance?: Average velocity = change in speed / change in time
|| || || || || || || || || || ||
60-.5-0 = || || ||
9.23 m/s^-1 ||
3. A soccer ball is kicked from the ground with an initial velocity of 27 m/s, at
|| || || || || || || || || || || || || || ||
an angle of 39 degrees.
|| || || || ||
a. What is the horizontal and vertical component at release?: 27 cos(39 = 20.98
| | | | | | | | || | | |
|| horizontal = 20.98 m/s^-1 || || ||
27 sin(39 = 16.99
|| || ||
Vertical = 16.99 m/s^-1 || || ||
4. A ball was launched off a cliff with an initial velocity of 15 m/s at an angle
|| || || || || || || || || || || || || || || ||
of 315 degrees ccw from the horizontal (45 degrees below the horizontal).
|| || || || || || || || || || || ||
a. What is the horizontal and vertical component at release?: 15 cos(315 =
|| || || || || || || || || || ||
10.60
||
horizontal = 10.60 m/s^-1 15 || || || ||
|| sin(315 = -10.60 || ||
1 ||/
||8
, BMC Test 2 Math || || ||
Study ||online ||at ||https://quizlet.com/_d1wvh9
Vertical = -10.60 m/s^-1 || || ||
5. A baseball is thrown with a velocity of 15ms-1 at an angle of 30 degrees ccw
| | | | | | | | | | | | | | |
from the horizontal.
|| || ||
a. What are the horizontal and vertical components of the velocity just as the
|| || || || || || || || || || || ||
ball leaves thehand?
|| || ||
b. What is the horizontal and vertical velocity of the ball when it reaches
|| || || || || || || || || || || ||
maximum height?
|| ||
c. At what time does the ball reach maximum vertical height?
|| || || || || || || || ||
d. How far has the ball traveled vertically and horizontally at this time?: a.
|| || || || || || || || || || || ||
15 sin(30
|| ||
7.5 m/s initial vertical || || ||
15 cos(30 ||
12.99 m/s initial horizontal || || ||
b.
horizontal = 12.99 m/s || || ||
vertical = 0 m/s
|| || || ||
c.
final velocity = initial velocity + (acceleration)(time) 0
|| || || || || || ||
= 7.5 + (-9.81)(t)
|| || || ||
-7.5/-9.81 = .7645 seconds || || ||
|| d.
vertical displacement = (initial vertical velocity)(time)+(1/2)(-9.81)(time^2) x
|| || || || || ||
= (7.5) (.7645) + (1/2) (-9.81) (.7645^2)
|| || || || || || ||
= 2.87 M || ||
Horizontal displacement = (initial horizontal velocity)(time) x || || || || || ||
= (12.99) (.7645)
|| || ||
= 9.94 M || ||
2 ||/
||8
