Worksheet: Acceleration Problems
Name
Solve the following problems by making a list, writing the equation, filling in the equation and then
solving the problem.
1. What is the average acceleration of a car driven by Bubba if the car goes from 22.0 miles/hour to 74.0
miles/hour in 8.56s?
List
Equation
Fill in equation
Work/Answer
a =?
vi = 22.0 mi/h
vf = 74.0 mi/h
t = 8.56s
a = vf - vi
a = 74mi/h – 22mi/h
t
52mi/h
8.56s
8.56s
a= 6.07mi/h/s
2. Billy Bob’s four-wheeler will accelerate at 3.0m/s/s. If Billy Bob starts at 5.00mi/h what will be his
final speed after 4.00s?
List
Equation
Fill in equation
Work/Answer
A = 3.0m/s/s
a = vf - vi
3.0m/s/s = Vf – 5mi/h
3.0m/s2(4.00s) = Vf – 5mi/h
vi = 5.00 mi/h
t
4.00s
12.0m/s = Vf – 5mi/h
vf = ?
12.0m/s + 5mi/h= Vf
t = 4.00s
12m/s + 2.2m/s = 14m/s
conversion: 5mi x 1 hour x 1600m = 2.22m/s
h
3600s
1 km
3. Darla Mae is going to school in her hometown of Murphy Mississippi and she is driving the family
Tractor. How long does it take Darla Mae to go from 0.00 mi/h to her max speed of 43mi/h if she
accelerates at a rate of 3.0mi/h/s?
List
Equation
Fill in equation
Work/Answer
A = 3.0mi/h/s a = vf - vi
3.0mi/h/s = 43mi/h – omi/h
3.0m/h· s = 43mi/h – 0mi/h
vi = 0.00 mi/h
t
t
t
vf = 43mi/h
t = 43mi/h – 0mi/h
t=?
3.0m/h· s
14mi/h
m/h· s
14s
4. Elroy’s crop duster plain must reach a speed of 120mi/h to take off. If Elroy starts at a rolling speed
of 5.00mi/h and accelerates at 7.55ft/s/s, how long must his take-off field be?
List
Equation
Fill in equation
Work/Answer
Vf = 120mi/h Vf2 = Vi2 + 2ad
Vi = 5mi/h
A = 7.55ft/s2
D =?
(120mi/h)2 = (5mi/h)2 - 2(7.55ft/s2)(d)
14400mi2/h2 = 25mi2/h2 + 15.1ft/s2(d)
14400mi2/h2 - 25mi2/h2 = 15.1ft/s2(d)
13375mi2/h2 = 15.1ft/s2(d)
13375mi2/h2 = d
15.1ft/s2
952mi2/h2 · s2/ft =d
D = 0.39mi or 2.0 x 103ft
5. Gertie Figbottom drops and egg from the hayloft in the barn. The egg falling under the influence of
gravity will take how long to hit the ground if the hayloft is 5.25 meters high?
List
Equation
Fill in equation
Work/Answer
D = 5.25 m
d = vi + at2
5.25m =(0)(t) + 9.81m/s2(t)2
5.25 = 9.81m/s2(t)2
A = 9.81m/s2
2
2
2
Vi= 0
2(5.25m) =9.81m/s2(t)2
10.50m = t2
9.81m/s2
1.07s2 =t2
√1.07s2 = t
1.03s =t
When I divided 9.81m/s2 into 10.5m the meters canceled out and the s2 went to the top
6. If Gertie’s egg hits her husband Claude in the head, what is the speed at which it hits his head?
Claude is 6.0ft and 2.0 inches tall. There are 2.54cm in 1.0inches.
List
Equation
Fill in equation
Work/Answer
Vf = ?
Vf2 = Vi2 + 2ad
Vf2 = (0)2 - 2(9.81m/s2)(3.37m)
Vi = 0
A = 9.81m/s2
Vf2 = 0 + 66.11m2/s2
Vf =√ 66.11m2/s2
Vf = 8.13m/s
D = 3.37m
To get the distance of 3.37m I subtracted 6ft 2in from 5.25m. First I converted 6ft 2in to inches only and
got 74 inches. Next I did the following 74 in x 2.54cm/in and got 188cm which became 1.88m. If you
subtract 1.88m from 5.25m you get 3.37m.
7. Jimbo’s suped-up camaro takes 0.0536mi to reach it max speed. If it will average an acceleration at
42.50mi/h/s, what will be his final speed if the initial speed is 0.00 mi/h?
List
Equation
Fill in equation
Work/Answer
Vf = ?
Vf2 = Vi2 + 2ad
Vf2 = (0)2 - 2(42.50mi/h·s)(0.0536mi)
Vi = 0
A =42.50mi/h·s
Vf2 = 0 + 4.556mi2/h·s
Vf =√ 16401mi2/h2
Vf = 128mi/h
D = 0.0536mi
I changed the acceleration unit from mi/h/s to mi/h·s because they are the same thing.
Conversion: I converted 4.556mi2/h·s to mi2/h2 by doing the following conversion
4.556mi2/h·s x 3600s/1hour = 16401mi2/h2
8. A car traveling at 120.0 Km/h comes upon Farmer Johnson’s prized 750lb Sow in the middle of the
road. If it takes the car 5.55 seconds to come to a complete stop how far from the Sow must the brakes
of the car be applied in order to stop 1.0 cm from the Sow? The rate of average rate acceleration is
6.00m/s/s. Note the 750 lbs is not needed, you must find how far the car will go in order to stop, then
you will need to add 1cm. You can do this one two ways.
List
Equation
Fill in equation
Work/Answer
Vi = 120km/h d = Vit + at2
d = 120km/h(5.55s) + (-6.00m/s2)(5.55s)2
d= 666km·s + (-184.815m)
T = 5.55s
2
2
h
2
A = 6.00m/s/s
or
d = 0.185Km + 92.41m
Vf2 = Vi2 + 2ad
O =(120km/h)2 +2(-6.00m/s2)d
d = 185m - 92.42m
D = 92.58m
So the answer is 92.59m