Chapter 3: Accelerated Motion

Chapter 3:Accelerated MotionHonors PhysicsGlencoe Science, 20053.1 AccelerationChange in time- ending time minus initial time Dt = tf - tiChange in velocity- final velocity minus initial velocity Dv = vf - viAverage acceleration- change in velocity between two distinct time intervals = Dv/Dt = (vf - vi)/(tf - ti)(m/s2)Instantaneous acceleration- change in velocity at an instant in timeOnly found by determining the tangent of a curve on a velocity-time graphMotion DiagramsShows an object slowing down, speeding up, without motion, or at constant motion

Velocity-Time GraphsRepresents motion graphicallyPlots the velocity versus the time of the object

3.2 Motion with Constant AccelerationSometimes we have an initial velocityLike when we pull through a stop light as it turns greenOur acceleration formula can be rearranged Dv= Dt vf - vi = DtIf we are looking for the final velocity, then we multiply the acceleration by the time and add the initial velocityvf = Dt + vi

Practice Problem 3.2.19A bus that is traveling at 30.0km/h speeds up at a constant rate of 3.5m/s2. What velocity does it reach 6.8s later?Convert all to terms of m and s:(30.0km/h)(1000m/km)(1h/3600s)=viDefine known & unknown:a=3.5m/s2vi=8.33m/s Dt=6.8svf=?Choose the appropriate equationvf=Dt + viRearrange if necessary (it is not in this case)Plug & Chugvf=(3.5m/s2)(6.8s)+(8.33m/s)vf=32.13m/sAnalyzing GraphsLooking at a position-time graph, we can find:Figure 3-9Velocity (slope)Specific positions at specific timesFrom our original velocity equation, v=Dd/Dt, we can find that Dd=vDtThe area under the line in a velocity-time graph is vDt, which is the displacement!Figure 3-10Slope is v/Dt, which is acceleration!Position with Constant AccelerationPosition (df) of an object under acceleration can be found with:df=at2(m/s2)(s2)=mIf there is an initial distance that we need to add, then:df=di+at2(m)+(m/s2)(s2)=mIf there is an initial velocity, then we can also include that term!df=di+vit+at2(m)+(m/s)(s)+(m/s2)(s2)=mThis equation is only useful if time of travel is known

Position with Constant AccelerationUnlike the prior equation, sometimes time is not known, so we need to relate velocity to distance traveledvf2=vi2+2a(df-di)(m/s)2=(m/s)2+(m/s2) (m-m)

Equations SummaryEquationVariablesInitial Conditionsvf = Dt + vivf, , Dtvidf=di+vit+at2df, t, adi, vivf2=vi2+2a(df-di)vf, a, dfvi, diPractice Problem 3.2.27aA 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:vi=44m/svf=22m/s Dt=11sDd=?a=?Choose the appropriate equationWe need to find firstvf=Dt + viRearrange if necessary=(-vi+vf)/DtPlug & Chug=(-44m/s+22m/s)/(11s)=-2m/s2Practice Problem 3.2.27bNow that we have , we can solve for dfDefine known & unknown:vi=44m/svf=22m/s Dt=11sa=-2m/s2 Dd=?Choose the appropriate equationdf=di+vit+at2Rearrange if necessaryPlug & Chugdf=(0m)+(44m/s)(11s)+(-2m/s2)(11s)2df=363mPractice Problem 3.2.28aA 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:vi=15m/svf=25m/s Dt=?a=? Dd=125mChoose the appropriate equationWe need to find first, but dont know timevf2=vi2+2a(df-di)Rearrange if necessarya=(vf2-vi2)/(2Dd)Plug & Chuga=((25m/s)2-(15m/s)2)/((2)(125m))a=1.6m/s2

Practice Problem 3.2.28bNow that we have , we can solve for DtDefine known & unknown:vi=15m/svf=25m/s Dt=?a=1.6m/s2 Dd=125mChoose the appropriate equationvf = Dt + viRearrange if necessary(vf-vi)/=DtPlug & Chug(25m/s-15m/s)/(1.6m/s2)=Dt Dt=6.3s3.3 Free FallFree Fall- an object falling only under the influence of gravityAcceleration due to gravity- an object speeds up due to the Earths gravitational pullg=9.8m/s2Gravity is a specific kind of acceleration: like a quarter is a specific kind of moneyGravity always points toward the center of the EarthFree Fall EquationsAs gravity (g) is a kind of acceleration (a), we can replace all of the as with gThis can only be done if the object is in free fallEquationVariablesInitial Conditionsvf =gDt + vivf, Dtvidf=di+vit+gt2df, tdi, vivf2=vi2+2g(df-di)vf, dfvi, diPractice Problem 3.3.42aA construction worker accidentally drops a brick from a high scaffold. What is the velocity of the brick after 4.0s?Define known & unknown:vi=0m/svf=? Dt=4.0sg=9.8m/s2 Dd=?Choose the appropriate equationvf =gDt + viRearrange if necessary (not necessary)Plug & Chugvf =(9.8m/s2)(4.0s)+(0m/s)vf =39.2m/sPractice Problem 3.3.42bA construction worker accidentally drops a brick from a high scaffold. How far does the brick fall during this time?Define known & unknown:vi=0m/svf=39.2m/s Dt=4.0sg=9.8m/s2 Dd=?Choose the appropriate equationdf=di+vit+gt2Rearrange if necessary (not necessary)Plug & Chugdf=(0m)+(0m/s)(4.0s)+(9.8m/s2)(4.0s)2df=78.4mPractice Problem 3.3.45aA 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:vi=22.5m/svf=0m/s Dt=?g=-9.8m/s2 Dd=?Choose the appropriate equationvf2=vi2+2g(df-di)Rearrange if necessary Dd=(vf2-vi2)/(2g)Plug & Chug Dd=((0m/s)2-(22.5m/s)2)/(2(-9.8m/s2)) Dd=25.82mPractice Problem 3.3.45bA 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:vi=22.5m/svf=0m/s Dt=?g=-9.8m/s2 Dd=25.82mChoose the appropriate equationvf =gDt + viRearrange if necessary Dt=(vf-vi)/gPlug & Chug Dt=(0m/s-22.5m/s)/(-9.8m/s2) Dt=2.30s

Time DilationAs an objects speed approaches 3.0x108m/s (c), the time as observed from the outside of the ship changesSo if you are travelling very fast,