How Far, or Linear Kinematics
by Chris Murray, November 1997
1.
What distance will a train stop in if its initial
velocity is 23 m/s and its
acceleration is -.25 m/s/s?
(1058 m)
Here is what you start with:
- s = ?
- t = ?
- u = 23 m/s
- va = ?
- v = 0 (it stops)
- a = -.25 m/s/s
Well, you can find time from v = u + at:
0 = 23 m/s + (-.25 m/s/s)t so t = 92 s
and you can find the displacement from s = t(u + v)/2
= 92*(23 + 0)/2 = 1058 m, the answer.
You also could use v2 = u2 + 2as and find the answer
in one step: 232 = 02 + 2(-.25 m/s/s)s so s = 1058 m
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2.
What distance will a car cover accelerating from 12 m/s to 26 m/s in 14 seconds?
(266 m)
Here is what you start with:
- s = ?
- t = 14 s
- u = 12 m/s
- va = ?
- v = 26 m/s
- a = ?
Find the displacement: s = t(u + v)/2
= 14*(12 + 26)/2 = 266 m
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3.
A person starts at rest and accelerates at 3.2 m/s/s for 3.0 seconds. What is their
final velocity? What is their average velocity? What distance do they cover in that
time?
(9.6 m/s, 4.8 m/s, 14.4 m)
Here is what you start with:
- s = ?
- t = 3.0 seconds
- u = 0 (They start at rest)
- va = ?
- v = ?
- a = 3.2 m/s/s
You can use v = u + at to find the final velocity: v = 0 + (3.2 m/s/s)(3.0 seconds) = 9.6 m/s
Then use va = (u + v)/2 to find the average velocity: va = (0 + 9.6 m/s)/2 = 4.8 m/s
and finally s = vat to find displacement: s = 4.8 m/s)(3.0 s) = 14.4 m
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4.
Steve Apt's group claimed that they fell 3.2 seconds from a cliff into the water.
What was their final speed? How high was the cliff? What is your favorite color?
(31.36 m/s, 50.2 m, Red)
Here is what you start with:
- s = ?
- t = 3.2 seconds
- u = 0 (We assume they don't hurl themselves down the cliff)
- va = ?
- v = ?
- a = 9.80 m/s/s (The acceleration of free fall on Earth)
Use v = u + at to find the final velocity: v = 0 + (9.80 m/s/s)(3.2 s) = 31.36 m/s
Then try 2as = v2 - u2 for finding the height of the cliff:
2(9.80 m/s/s)s = 31.36 m/s2 - 02 = 983.4496 m2/s2
so s = 983.4496 m2/s2/19.6 m/s/s = 50.176 m or about 50.2 m (sf 50. m)
As far as your favorite color, I guess that is a matter of personal choice. I think
that red is a fine personal choice, so I chose it. Don't you think you ought to
choose red too?
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5.
A car going 12.7 m/s accelerates for 14 seconds at 2.6 m/s/s. What is its final
velocity? What distance does it go during that time?
(49.1 m/s, 432.6 m)
Here is what you start with:
- s = ?
- t = 14 seconds
- u = 12.7 m/s
- va = ?
- v = ?
- a = 2.6 m/s/s
Use v = u + at to find the final velocity: v = 12.7 m/s + (2.6 m/s/s)(14 s) = 49.1 m/s
then find the average velocity using va = (u + v)/2: va = (12.7 m/s + 49.1 m/s)/2 = 30.9 m/s
finally, use s = vat for the displacement: s = 30.9 m/s)(14 seconds) = 432.6 m or about 430 m with sig figs.
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6.
What time will it take you to hit the water off of a 10.0 m board? What speed will you
be going when you hit the water?
(1.43 s, 14 m/s)
Here is what you start with:
- s = 10.0 m (10.0 m boards are that high, not that long)
- t = ?
- u = 0 (assuming falling from rest)
- va = ?
- v = ?
- a = 9.80 m/s/s (gravity)
to find the time, you need to use s = ut + 1/2at2:
10.0 m = 0t + 1/2(9.80 m/s/s)t2
doing algebra,
2(10.0 m)/(9.80 m/s/s) = t2
so you get an absolute value of time of:
t = 1.4286 s.
then to find the final velocity, the easiest way is to use v = u + at: v = 0 + (9.80 m/s/s)(1.4286 s) = 14 m/s
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7.
A car slows from 42 m/s to 21 m/s over a distance of 84 m. What was the
average velocity? What was the time? What was the acceleration?
(31.5 m/s, 2.67 s, -7.9 m/s/s)
Here is what you start with:
- s = 84 m
- t = ?
- u = 42 m/s
- va = ?
- v = 21 m/s
- a = ?
Find the average velocity with va = (u + v)/2: va = (42 m/s + 21 m/s)/2 = 31.5 m/s
Then find the time from s = vat
84 m = (31.5 m/s)t
so t = 84 m / 31.5 m/s = 2.6667 s
and finally, v = u + at will find the acceleration:
21 m/s = 42 m/s + a(2.6667 s)
so
-21 m/s = a(2.6667 s), and a = -21 m/s /(2.6667 s) = -7.875 m/s/s (-7.9 m/s/s with sf)
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8.
A car accelerates from rest down a hill reaching a final speed of 13.7 m/s
over a
distance of 56 m. What was the average speed? What was the time? What was the
acceleration?
(6.85 m/s, 8.2 s, 1.68 m/s/s)
Here is what you start with:
- s = 56 m
- t = ?
- u = 0 (from rest)
- va = ?
- v = 13.7 m/s
- a = ?
Find the average velocity: va = (u + v)/2 = (0 + 13.7 m/s)/2 = 6.85 m/s
Find time: s = vat plugging in numbers:
56 m = (6.85 m/s)t, so t = 56 m / 6.85 m/s = 8.1752 s
Use v = u + at to find the acceleration:
13.7 m/s = 0 + a(8.1752 s), so a = 13.7 m/s / 8.1752 s = 1.6758 m/s/s (1.7 m/s/s w sf)
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9.
A car skids to a halt in 34 m with an acceleration of 8.2 m/s/s. What was the initial
velocity?
(23.6 m/s)
Here is what you start with:
- s = 34 m
- t = ?
- u = ?
- va = ?
- v = 0 (to a halt)
- a = -8.2 m/s/s (It must be negative if the 34 m is positive)
Police investigating accident scenes solve this problem to determine fault in the case
of excessive speeding. The 34 m would be the length of the skid marks, and the tires
and car would have a characteristic acceleration.
I would use 2as = v2 - u2 to find the answer in one swell foop:
2(-8.2 m/s/s)(34 m) = 02 - u2 = -u2
getting rid of the minus sign on both sides, and combining on the left side:
557.6 m2/s2 = u2, so the absolute value of u must be:
23.61 m/s. Only the positive value has meaning. (This is about 53 mph - Zippy for a
residential neighborhood)
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10.
What must be the acceleration of a train in order for it to stop from 12 m/s in a
distance of 541 m?
(-.13 m/s/s)
Here is what you start with:
- s = 541 m
- t = ?
- u = 12 m/s
- va = ?
- v = 0 (stop)
- a = ?
Use 2as = v2 - u2:
2a(541 m) = 02 - (12 m/s)2 = -144 m2/s2
so
(1082 m)a = -144 m2/s2, a = -144 m2/s2/(1082 m) = -.1331 m/s/s or -.13 m/s/s
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