Section 1
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the instantaneous acceleration at any point is equal to the
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Date created
Mar 1, 2020
Cards (69)
Section 1
(50 cards)
the instantaneous acceleration at any point is equal to the
average acceleration over the entire time interval
equations
*
Speed
--the distance traveled per unit of time -- unit of velocity -- v = d/t
Average speed is a
function of total distance and total time
the common force every object on earth experiences
gravity
v = sqr root
2ax
average acceleration 2
vf-vi/tf-ti
vf = vo + at
2
Displacement, Delta X
-- the straight line separation of two points in a specified direction --contains magnitude AND direction, a number, unit and angle -- Positive or negative based on location
position vs time graph
--(d-t) or (x-t) graphs -- slope exact value in m/s -- positive slope = forward -- negative slope = down -- horizontal line = rest (0)
velocity equation
v = vo + at
An automobile with an initial speed of 4.30 m/s accelerates uniformly at a rate of 3.00 m/s2. Find the final speed and the displacement after 5.00 s.
-- vi = 4.30, a = 3.00, Δt = 5.00, Δx = ?, vf = ? -- solve for vf by using equation 4 -- solve for Δx by using equation 3
A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s
a = -9.8, t = 2, 4, 7, vo = + 30, y =?, v = ? -- y = (30)t + .5 (-9.8)t^2 -- find v from the equitation that contains v and no x -- substitute t v = 10.4, 9.20, 38.6 m/s^2
the acceleration is proportional to the
magnitude of the force
Signs for displacement
-- right of origin is positive -- left of origin is negative
Distance d
-- the length of the actual path taken by an object --scalar quantity --Point A to Point B
velocity vs time graph
- v-t graph -- slope = rise/run = change in velocity/change in time = acceleration -- delta y times delta x is equal to delta v times delta t, which is equal to distance -- area under the curve gives the distance traveled (how much you've changed) --horizontal line = constant speed -- whenever you cross the x-axis, you change direction
average speed equals
total path m / time
acceleration of gravity
-9.81 m/s^2 or -32 ft/s^2
xf = xo + .5(vf + vo)t
3
average velocity 1
net displacement (delta x)/ time m/s
If a person runs forward at 5 m/s and then throws a ball at 3 m/s in the same direction they are running, then how fast is the ball going relative to the ground?
5 m/s + 3 m/s = 8 m/s forward
gravity is always
negative
equation 2
--average acceleration --vf = vi + aΔt
average velocity equations
xf-xi/tf-ti vi+vf/2
Velocity
--the displacement per unit of time -- v = d/t = m/s -- + or - based on direction of motion
equation 4
--final velocity -- vf^2=vi^2+2aΔx
displacement equation
vt
converting graphs: displacement (m) to velocity (m/s)
-- find the slope -- slope = y-axis number -- linear lines in d-t graph is a horizontal line on v-t -- 0 displacement/horizontal line = horizontal line at 0 -- curve = constant slope --* CURVE NEEDS TANGENTIAL LINE --when drawing, just draw the lines first, then connect the dots
instantaneous velocity
the slope of the tangential line at that point
acceleration
the change in velocity per unit of time --vector quantity --speeding up --slowing down --turning
converting graphs: v-t to d-t
-- split the graph lines --take the area -- linear line of area = a curve on d-t graph, levels out at y-axis (slope of v-t) --horizontal on v-t =linear line on d-t (go from previous point straight down to -16) --horizontal line on x-axis is constant speed on d-t (horizontal line)
v^2 =
2ax
converting graphs: v-t to a-t
-- use slope --horizontal lines on the v-t = horizontal lines on x-axis -- linear lines slope = y-axis number on a-t **REMEMBER TO GRAPH LINES FIRST, CONNECT DOTS LATER
the average speed depends ONLY on the
distance traveled and the time required
v = 0
no sign needed
v equation
v = change in x / change in t
the direction of acceleration
is the same as the direction of force -- a will always have the same sign of F, regardless of velocity
If a person runs forward at 5 m/s and then throws BACKWARDS ball at 3 m/s in the same direction they are running, then how fast is the ball going relative to the ground?
5 m/s + 3 m/s = 2 m/s
t =
v / a
equation 1
--average velocity --Δx = .5(vi+vf)Δt
xf = xo + vot + .5at^2
1
a equation
a = change in v / change in t
initial velocity
vo
converting graphs: a-t to v-t
--horizontal lines on a-t graph = linear line on v-t --**y-intercept where horizontal line is at is the slope of linear line
equation 3
-- displacement -- Δx = viΔt + .5a(Δt)^2
average acceleration 1
vf-vo/tf-to
A runner runs 200 m east, then changes direction and runs 300 m west. If the entire trip takes 60 s, what is the velocity?
8.33 m/s
frame of reference
the speed of every object is relative to something else
a change in velocity requires the application of
force
Section 2
(19 cards)