Problem #1: A skater goes from a standstill to a speed of

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Transcript Problem #1: A skater goes from a standstill to a speed of

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 = 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
Answer
• 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
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
Answer
• a= 2(Δx - vi2Δt)…plug in values
(Δt)2
• a= 2(640m – 8m/s·40s)
(40s)2
• a = 0.4 m/s2
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
Answer
• Use vf=aΔt, vf= (5m/s2)(4s)=20 m/s
• Δx=1/2a(Δt)2 , Δx=1/2(5m/s2)(4s)2=40m
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
Answer
• vf=aΔt, vf= (9.81m/s2)(5s)= 49.1m/s
• Δx=1/2a(Δt)2 = ½(9.81m/s2)(5s)2= 123m
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
Answer
• a= vf2
2Δx
• a=(80ft/s)2
2(250ft)
• a=13ft/s2
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
Answer
• (vf)2=2aΔx
• Solve for vf, vf= √(2aΔx)
• vf= √{2(0.30m/s2)(50.m)} =5.5m/s
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)2 We call
displacement y in this problem because the motion is up
and down, the direction of the y axis.
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
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
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)