# How Far, or Linear Kinematics

by Chris
Murray, November 1997, Derek Clarno and Chris Murray 2003
## 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:
- x = ?
- Dt = ?
- v
_{i} = 23 m/s
- v
_{f} = 0 (it stops)
- a = -.25 m/s/s

Well, you can find time from v_{f} = v_{i} + a(Dt): 0 = 23 m/s
+ (-.25 m/s/s)t so t = 92 s and you can find the displacement from Dx =
^{1}/_{2}(v_{i} + v_{f})Dt_{ }= 92*(23 + 0)/2 = 1058 m, the answer. You also could use v_{f}^{2} =
v_{i}^{2} + 2a(Dx) and find the answer in one step:
23^{2} = 0^{2} + 2(-.25 m/s/s)x so x = 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:
- x = ?
- Dt = 14 s
- v
_{i} = 12 m/s
- v
_{f} = 26 m/s
- a = ?

Find the displacement: Dx =
^{1}/_{2}(v_{i} + v_{f})Dt_{ }= ^{1}/_{2}(12 + 26)(14) = 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:
- x = ?
- Dt = 3.0 seconds
- v
_{i} = 0 (They start at rest)
- v
_{f} = ?
- a = 3.2 m/s/s

You can use v_{f} = v_{i} + a(Dt) to find the final velocity: v
= 0 + (3.2 m/s/s)(3.0 seconds) = 9.6 m/s The average of 0 and 9.6 m/s is 4.8 m/s
(9.6 + 0)/2 and finally Dx =
^{1}/_{2}(v_{i} + v_{f})Dt to find displacement: x = ^{1}/_{2}(0 +
9.6 m/s)Dt(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:
- x = ?
- Dt = 3.2 seconds
- v
_{i} = 0 (We assume they don't hurl themselves down the cliff)
- v
_{f} = ?
- a = 9.80 m/s/s (The acceleration of free fall on Earth)

Use v_{f} = v_{i} + a(Dt) to find the final velocity: v = 0 + (9.80 m/s/s)(3.2 s) = 31.36 m/s Then
try v_{f}^{2} =
v_{i}^{2} + 2a(Dx) for finding the height of the cliff:
(31.36 m/s)^{2} = 0^{2} + 2(9.80 m/s/s)x, x = 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:
- x = ?
- Dt = 14 seconds
- v
_{i} = 12.7 m/s
- v
_{f} = ?
- a = 2.6 m/s/s

Use v_{f} = v_{i} + a(Dt) to find the final velocity: v = 12.7
m/s + (2.6 m/s/s)(14 s) = 49.1 m/s then find the displacement using x = ^{1}/_{2}(v_{i} + v_{f})Dt = ^{1}/_{2}(12.7 m/s + 49.1 m/s)(14 s) =
432.6 m
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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:
- x = 10.0 m (10.0 m boards are that high, not that long)
- Dt = ?
- v
_{i} = 0 (assuming falling from rest)
- v
_{f} = ?
- a = 9.80 m/s/s (gravity)

to find the time, you need to use s = ut
+ ^{1}/_{2}at^{2}: 10.0 m = 0t +
^{1}/_{2}(9.80 m/s/s)t^{2} doing algebra, 2(10.0
m)/(9.80 m/s/s) = t^{2} 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_{f} = v_{i} + a(Dt):
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:2.6666
- x = 84 m
- Dt = ?
- v
_{i} = 42 m/s
- v
_{f} = 21 m/s
- a = ?

First let's find the acceleration using v_{f}^{2}
= v_{i}^{2} + 2a(Dx): 21^{2}
= 42^{2} + 2(a)84, a = -7.875 m/s/s, find the time using v_{f} =
v_{i} + a(Dt): 21 = 42 + (-7.875)t, t = 2.667
s. The average of 21 and 42 is 31.5 m/s (21+42)/2 = 31.5
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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:
- x = 56 m
- Dt = ?
- v
_{i} = 0 (from rest)
- v
_{f} = 13.7 m/s
- a = ?

Find the average velocity: v (average) =
^{1}/_{2}(v_{i} + v_{f}) = (0 + 13.7 m/s)/_{2} = 6.85 m/s Find time:
Dx = ^{1}/_{2}(v_{i} + v_{f})Dt
plugging in numbers: 56 m = ^{1}/_{2}(0 + 13.7)t, so t =
8.1752 s Use v_{f} = v_{i} + a(Dt) 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
- Dt = ?
- v
_{i} = ?
- v
_{f} = 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 v_{f}^{2} =
v_{i}^{2} + 2a(Dx) to find the answer in one swell
foop: 0^{2 }=^{ }
v_{i}^{2} + 2(-8.2 m/s/s)(34 m). Getting rid of the minus sign on both sides, and combining on the left side:
557.6 m^{2}/s^{2} =
v_{i}^{2}, so the absolute value of
v_{i} 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:
- x = 541 m
- Dt = ?
- v
_{i} = 12 m/s
- v
_{f} = 0 (stop)
- a = ?

Use v_{f}^{2} =
v_{i}^{2} + 2a(Dx):
0^{2} = (12 m/s)^{2 }+^{ }2a(541 m), a = -144 m^{2}/s^{2}, a = -144
m^{2}/s^{2}/(1082 m) = -.1331 m/s/s or -.13 m/s/s
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