Honors Physics Glencoe Science, 2005

 Change in time- ending time minus initial time  t = t f - t i  Change in velocity- final velocity minus initial velocity  v = v f - v i  Average acceleration- change in velocity between two distinct time intervals ā =  v/  t = (v f - v i )/(t f - t i )(m/s 2 )  Instantaneous acceleration- change in velocity at an instant in time Only found by determining the tangent of a curve on a velocity-time graph

 Shows an object slowing down, speeding up, without motion, or at constant motion

 Represents motion graphically  Plots the velocity versus the time of the object

 Sometimes we have an initial velocity Like when we pull through a stop light as it turns green  Our acceleration formula can be rearranged  v= ā  t  v f - v i = ā  t  If we are looking for the final velocity, then we multiply the acceleration by the time and add the initial velocity v f = ā  t + v i

 A bus that is traveling at 30.0km/h speeds up at a constant rate of 3.5m/s 2. What velocity does it reach 6.8s later?  Convert all to terms of m and s: (30.0km/h)(1000m/km)(1h/3600s)=v i  Define known & unknown: a=3.5m/s 2 v i =8.33m/s  t=6.8sv f =?  Choose the appropriate equation v f = ā  t + v i  Rearrange if necessary (it is not in this case)  Plug & Chug v f =(3.5m/s 2 )(6.8s)+(8.33m/s) v f =32.13m/s

 Looking at a position-time graph, we can find: Figure 3-9 Velocity (slope) Specific positions at specific times  From our original velocity equation, v=  d/  t, we can find that  d=v  t  The area under the line in a velocity-time graph is v  t, which is the displacement! Figure 3-10 Slope is v/  t, which is acceleration!

 Position (d f ) of an object under acceleration can be found with: d f =½at 2 (m/s 2 )(s 2 )=m  If there is an initial distance that we need to add, then: d f =d i +½at 2 (m)+(m/s 2 )(s 2 )=m  If there is an initial velocity, then we can also include that term! d f =d i +v i t+½at 2 (m)+(m/s)(s)+(m/s 2 )(s 2 )=m This equation is only useful if time of travel is known

 Unlike the prior equation, sometimes time is not known, so we need to relate velocity to distance traveled v f 2 =v i 2 +2a(d f -d i )(m/s) 2 =(m/s) 2 +(m/s 2 ) (m-m)

EquationVariablesInitial Conditions v f = ā  t + v i v f, ā,  t vivi d f =d i +v i t+½at 2 d f, t, ad i, v i v f 2 =v i 2 +2a(d f -d i )v f, a, d f v i, d i

 A race car travels on a racetrack at 44m/s and slows at a constant rate to a velocity of 22m/s over 11s. How far does it move during this time?  Define known & unknown: v i =44m/sv f =22m/s  t=11s  d=? a=?  Choose the appropriate equation We need to find ā first v f = ā  t + v i  Rearrange if necessary ā =(-v i +v f )/  t  Plug & Chug ā =(-44m/s+22m/s)/(11s) ā =-2m/s 2

 Now that we have ā, we can solve for d f  Define known & unknown: v i =44m/sv f =22m/s  t=11sa=-2m/s 2  d=?  Choose the appropriate equation d f =d i +v i t+½at 2  Rearrange if necessary  Plug & Chug d f =(0m)+(44m/s)(11s)+½(-2m/s 2 )(11s) 2 d f =363m

 A car accelerates at a constant rate from 15m/s to 25m/s while it travels a distance of 125m. How long does it take to achieve this speed?  Define known & unknown: v i =15m/sv f =25m/s  t=?a=?  d=125m  Choose the appropriate equation We need to find ā first, but don’t know time v f 2 =v i 2 +2a(d f -d i )  Rearrange if necessary a=(v f 2 -v i 2 )/(2  d)  Plug & Chug a=((25m/s) 2 -(15m/s) 2 )/((2)(125m)) a=1.6m/s 2

 Now that we have ā, we can solve for  t  Define known & unknown: v i =15m/sv f =25m/s  t=?a=1.6m/s 2  d=125m  Choose the appropriate equation v f = ā  t + v i  Rearrange if necessary (v f -v i )/ ā =  t  Plug & Chug (25m/s-15m/s)/(1.6m/s 2 )=  t  t=6.3s

 Free Fall- an object falling only under the influence of gravity  Acceleration due to gravity- an object speeds up due to the Earth’s gravitational pull g=9.8m/s 2 Gravity is a specific kind of acceleration: like a quarter is a specific kind of money  Gravity always points toward the center of the Earth

 As gravity (g) is a kind of acceleration (a), we can replace all of the “a”’s with “g”  This can only be done if the object is in free fall EquationVariablesInitial Conditions v f =g  t + v i v f,  t vivi d f =d i +v i t+½gt 2 d f, td i, v i v f 2 =v i 2 +2g(d f -d i )v f, d f v i, d i

 A construction worker accidentally drops a brick from a high scaffold. What is the velocity of the brick after 4.0s?  Define known & unknown: v i =0m/sv f =?  t=4.0sg=9.8m/s 2  d=?  Choose the appropriate equation v f =g  t + v i  Rearrange if necessary (not necessary)  Plug & Chug v f =(9.8m/s 2 )(4.0s)+(0m/s) v f =39.2m/s

 A construction worker accidentally drops a brick from a high scaffold. How far does the brick fall during this time?  Define known & unknown: v i =0m/sv f =39.2m/s  t=4.0sg=9.8m/s 2  d=?  Choose the appropriate equation d f =d i +v i t+½gt 2  Rearrange if necessary (not necessary)  Plug & Chug d f =(0m)+(0m/s)(4.0s)+½(9.8m/s 2 )(4.0s) 2 d f =78.4m

 A tennis ball is thrown straight up with an initial speed of 22.5m/s. It is caught at the same distance above the ground. How high does the ball rise?  Define known & unknown: v i =22.5m/sv f =0m/s  t=?g=-9.8m/s 2  d=?  Choose the appropriate equation v f 2 =v i 2 +2g(d f -d i )  Rearrange if necessary  d=(v f 2 -v i 2 )/(2g)  Plug & Chug  d=((0m/s) 2 -(22.5m/s) 2 )/(2(-9.8m/s 2 ))  d=25.82m

 A tennis ball is thrown straight up with an initial speed of 22.5m/s. It is caught at the same distance above the ground. How long does the ball remain in the air?  Define known & unknown: v i =22.5m/sv f =0m/s  t=?g=-9.8m/s 2  d=25.82m  Choose the appropriate equation v f =g  t + v i  Rearrange if necessary  t=(v f -v i )/g  Plug & Chug  t=(0m/s-22.5m/s)/(-9.8m/s 2 )  t=2.30s

 As an objects’ speed approaches 3.0x10 8 m/s (c), the time as observed from the outside of the ship changes  So if you are travelling very fast,