# 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 = ?
• vi = 23 m/s
• vf = 0 (it stops)
• a = -.25 m/s/s
Well, you can find time from vf = vi + 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(vi + vf)Dt = 92*(23 + 0)/2 =  1058 m, the answer. You also could use vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 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
• vi = 12 m/s
• vf = 26 m/s
• a = ?
Find the displacement: Dx = 1/2(vi + vf)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
• vi = 0 (They start at rest)
• vf = ?
• a = 3.2 m/s/s
You can use vf = vi + 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(vi + vf)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
• vi = 0 (We assume they don't hurl themselves down the cliff)
• vf = ?
• a = 9.80 m/s/s (The acceleration of free fall on Earth)
Use vf = vi + a(Dt) to find the final velocity: v = 0 + (9.80 m/s/s)(3.2 s) = 31.36 m/s Then try vf2 = vi2 + 2a(Dx) for finding the height of the cliff: (31.36 m/s)2 = 02 + 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
• vi = 12.7 m/s
• vf = ?
• a = 2.6 m/s/s
Use vf = vi + 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(vi + vf)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 = ?
• vi = 0 (assuming falling from rest)
• vf = ?
• 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 vf = vi + 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 = ?
• vi = 42 m/s
• vf = 21 m/s
• a = ?
First let's find the acceleration using vf2 = vi2 + 2a(Dx): 212 = 422 + 2(a)84, a = -7.875 m/s/s, find the time using vf = vi + 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 = ?
• vi = 0 (from rest)
• vf = 13.7 m/s
• a = ?
Find the average velocity: v (average) = 1/2(vi + vf) = (0 + 13.7 m/s)/2 = 6.85 m/s Find time: Dx = 1/2(vi + vf)Dt plugging in numbers: 56 m = 1/2(0 + 13.7)t, so t  = 8.1752 s Use vf = vi + 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 = ?
• vi = ?
• vf = 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 vf2 = vi2 + 2a(Dx) to find the answer in one swell foop: 02 = vi2 + 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 m2/s2 = vi2, so the absolute value of vi 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 = ?
• vi = 12 m/s
• vf = 0 (stop)
• a = ?
Use vf2 = vi2 + 2a(Dx): 02 = (12 m/s)2 + 2a(541 m), a = -144 m2/s2, a = -144 m2/s2/(1082 m) = -.1331 m/s/s or -.13 m/s/s

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