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Problem #1:   A skater goes from a standstill to a speed of 6.7 m/s in 12 seconds.  What is the acceleration of the ska

Chapter 2 Review. Problem #1:   A skater goes from a standstill to a speed of 6.7 m/s in 12 seconds.  What is the acceleration of the skater?. Step 1:   Write down the equation needed for solving for acceleration. a =  v f – v i  =   Δ v             t              t

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Problem #1:   A skater goes from a standstill to a speed of 6.7 m/s in 12 seconds.  What is the acceleration of the ska

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  1. Chapter 2 Review Problem #1:   A skater goes from a standstill to a speed of 6.7 m/s in 12 seconds.  What is the acceleration of the skater?

  2. Step 1:   Write down the equation needed for solving for acceleration. • a =  vf – vi =   Δv             t              t • Step 2:   Insert the known measurements into the equation.Known :  The initial speed of the skater was zero since he was not in motion.  The skater finally reached a speed of 6.7m/s in 12 seconds, which is the final speed or velocity.  The equation will look like this: a = 6.7m/s – 0m/s = 6.7m/s =                 12s              12s

  3. Step 3:   Solve.  Carefully put all measurements into your calculator. You must solve the change in velocity portion of the equation before you can do the division portion to solve for acceleration.  Don't forget that the SI unit for acceleration is m/s2  . SOLUTION:The skater had an acceleration of  .56m/s2 . a = 6.7m/s – 0m/s = 12s 6.7m/s = .56m/s2   12s             Answer

  4. Problem #2 • A body with an initial velocity of 8 m/s moves with a constant acceleration and travels 640 m in 40 seconds. Find its acceleration. • You are given intial velocity (vi), displacement (Δx), and total time(Δt), But not final velocity…choose your equations • Your right!! Δx= vi2Δt +1/2a(Δt)2 • Solve for a: a= 2(Δx - vi2Δt)…plug in values (Δt)2

  5. Answer • a= 2(Δx - vi2Δt)…plug in values (Δt)2 • a= 2(640m – 8m/s·40s) (40s)2 • a = 0.4 m/s2

  6. Problem #3 • A car has a uniformly accelerated motion of 5 m/s2. Find the speed acquired and distance traveled in 4 seconds from rest. • You know that velocity initial (vi) = 0m/s, a=5 m/s2, and total time = 4s. You need to find final velocity (vf) and displacement (Δx) • Choose your equation…. • Your right!!!... Use vf=aΔt then, Δx=1/2a(Δt)2

  7. Answer • Use vf=aΔt, vf= (5m/s2)(4s)=20 m/s • Δx=1/2a(Δt)2, Δx=1/2(5m/s2)(4s)2=40m

  8. Problem #4 • A marble is dropped from a bridge and strikes the water in 5 seconds. Calculate the speed with which it strikes and the height of the bridge. • You are given total time(Δt)= 5s, velocity initial (vi) = 0m/s, and acceleration due to gravity a=g=9.81m/s2. You are to find velocity final (vf) and displacement Δx • Choose your equation…. • Your right!!! Use vf=aΔt then, Δx=1/2a(Δt)2

  9. Answer • vf=aΔt, vf= (9.81m/s2)(5s)= 49.1m/s • Δx=1/2a(Δt)2= ½(9.81m/s2)(5s)2= 123m

  10. Problem #5 • A car starts from rest and accelerates uniformly to a velocity of 80 ft/s after traveling 250 ft. Find its acceleration. • You are given velocity initial (vi) = 0m/s, velocity final (vf)= 80 ft/s (what is this???…we don’t work in ft/s, it’s fine because your distance is given in ft. You can express your acceleration in ft/s2. The equations do not care what units you use…they will work just the same.) And displacement Δx= 250 ft • Choose you equation……. • Your right!!! vf2= 2aΔx, solve for a • a= vf2 2Δx

  11. Answer • a= vf2 2Δx • a=(80ft/s)2 2(250ft) • a=13ft/s2

  12. Problem #6 • What velocity is attained by an object which is accelerated at 0.30 m/s2 for a distance of 50. m with a starting velocity of 0.0 m/s? • You are given acceleration, displacement, initial velocity…you are to find final velocity (vf)… • Choose your equation….. • Your right!!! (vf)2=2aΔx

  13. Answer • (vf)2=2aΔx • Solve for vf, vf= √(2aΔx) • vf= √{2(0.30m/s2)(50.m)} =5.5m/s

  14. Problem #7 • If a calculator falls off the side of student desk and hits the floor 0.39 seconds later, how tall is the desk and how fast would the calculator be traveling when it hits the floor? • You are given total time (Δt) = 0.39s, initial velocity (vi) = 0m/s, and acceleration due to gravity a=g=9.81m/s2, you are to find displacement (Δy), and final velocity (vf) • Choose your equation….. • Your right!!!! vf=aΔt and Δy=1/2a(Δt)2We call displacement y in this problem because the motion is up and down, the direction of the y axis.

  15. Answer • vf=aΔt = (9.81m/s2)(0.39s)= 3.8m/s • Δy=1/2a(Δt)2= ½ (9.81m/s2)(0.39s)2 • Δy= 0.75m

  16. Problem #8 • If a rocket in space is moving at a constant velocity of 9.8 m/s and then uses its propulsion system to accelerate to 10. m/s during a 3.0 minute burn, what would be the acceleration of the rocket? • Your are given initial velocity (vi)=9.8m/s, final velocity (vf) =10.m/s, and total time (Δt)=3.0min. We will need to convert this value into seconds to be used with velocities expressed in m/s. We need to find acceleration(a). • Choose your equation…. • Your right!!!! vf = vi + aΔt

  17. Answer • vf = vi + aΔt • Solve for a • a= vf - vi Δt Convert 3.0min to seconds 3.0min 60s = 180s 1min • a=(10.m/s-9.8m/s) = 1.1x10-3m/s2 (180s-0s)

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