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Honors Physics Glencoe Science, 2005 Change in • Dt = tf - ti Change in velocity time- ending time minus initial time velocity- final velocity minus initial • Dv = vf - vi Average acceleration- change in velocity between two distinct time intervals • ā = Dv/Dt = (vf - vi)/(tf - ti) Instantaneous (m/s2) 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 • Dv= ā Dt vf - vi = ā Dt If we are looking for the final velocity, then we multiply the acceleration by the time and add the initial velocity • vf = ā Dt + vi A 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)=vi Define known & unknown: • a=3.5m/s2 • Dt=6.8s vi=8.33m/s vf=? Choose the appropriate equation • vf=āDt + vi Rearrange if necessary (it is not in this case) Plug & Chug • vf=(3.5m/s2)(6.8s)+(8.33m/s) • vf=32.13m/s Looking find: at a position-time graph, we can • Figure 3-9 • Velocity (slope) • Specific positions at specific times From our original velocity equation, v=Dd/Dt, we can find that Dd=vDt The area under the line in a velocity-time graph is vDt, which is the displacement! • Figure 3-10 • Slope is v/Dt, which is acceleration! Position (df) of an object under acceleration can be found with: • df=½at2 (m/s2)(s2)=m • df=di+½at2 (m)+(m/s2)(s2)=m If there is an initial distance that we need to add, then: If there is an initial velocity, then we can also include that term! • df=di+vit+½at2 (m)+(m/s)(s)+(m/s2)(s2)=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 • vf2=vi2+2a(df-di) (m/s)2=(m/s)2+(m/s2) (m-m) Equation Variables Initial Conditions vf = ā Dt + vi vf, ā, Dt vi df=di+vit+½at2 df, t, a di, vi vf2=vi2+2a(df-di) vf, a, df vi, di 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: • vi=44m/s • Dt=11s • a=? vf=22m/s Dd=? Choose the appropriate equation • We need to find ā first • vf=āDt + vi Rearrange if necessary • ā=(-vi+vf)/Dt Plug & Chug • ā=(-44m/s+22m/s)/(11s) • ā=-2m/s2 that we have ā, we can solve for df Define known & unknown: Now • vi=44m/s • Dt=11s • Dd=? vf=22m/s a=-2m/s2 Choose the appropriate • df=di+vit+½at2 Rearrange if necessary equation Plug & Chug • df=(0m)+(44m/s)(11s)+½(-2m/s2)(11s)2 • df=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: • vi=15m/s • Dt=? • Dd=125m vf=25m/s a=? Choose the appropriate equation • We need to find ā first, but don’t know time • vf2=vi2+2a(df-di) Rearrange if necessary • a=(vf2-vi2)/(2Dd) Plug & Chug • a=((25m/s)2-(15m/s)2)/((2)(125m)) • a=1.6m/s2 that we have ā, we can solve for Dt Define known & unknown: Now • vi=15m/s • Dt=? • Dd=125m vf=25m/s a=1.6m/s2 Choose the appropriate equation • vf = ā Dt + vi Rearrange if necessary • (vf-vi)/ā=Dt Plug & Chug • (25m/s-15m/s)/(1.6m/s2)=Dt • Dt=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/s2 • 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 Equation Variables Initial Conditions vf =gDt + vi vf, Dt vi df=di+vit+½gt2 df , t di, vi vf2=vi2+2g(df-di) vf, df vi, di A 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/s • Dt=4.0s • Dd=? vf=? g=9.8m/s2 Choose the appropriate equation • vf =gDt + vi Rearrange if necessary (not necessary) Plug & Chug • vf =(9.8m/s2)(4.0s)+(0m/s) • vf =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: • vi=0m/s • Dt=4.0s • Dd=? vf=39.2m/s g=9.8m/s2 Choose the appropriate equation • df=di+vit+½gt2 Rearrange if necessary (not necessary) Plug & Chug • df=(0m)+(0m/s)(4.0s)+½(9.8m/s2)(4.0s)2 • df=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: • vi=22.5m/s • Dt=? • Dd=? vf=0m/s g=-9.8m/s2 Choose the appropriate equation • vf2=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=25.82m 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: • vi=22.5m/s • Dt=? • Dd=25.82m vf=0m/s g=-9.8m/s2 Choose the appropriate equation • vf =gDt + vi Rearrange if necessary • Dt=(vf-vi)/g Plug & Chug • Dt=(0m/s-22.5m/s)/(-9.8m/s2) • Dt=2.30s As an objects’ speed approaches 3.0x108m/s (c), the time as observed from the outside of the ship changes So if you are travelling very fast,