How Far, or Linear Kinematics

by Chris Murray, November 1997

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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: 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 Go to: Problem Formulas Table of Contents

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: Find the displacement: s = t(u + v)/2 = 14*(12 + 26)/2 =  266 m Go to: Problem Formulas Table of Contents

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: 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 Go to: Problem Formulas Table of Contents

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: 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? Go to: Problem Formulas Table of Contents

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: 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. Go to: Problem Formulas Table of Contents

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: 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 Go to: Problem Formulas Table of Contents

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: 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) Go to: Problem Formulas Table of Contents

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: 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) Go to: Problem Formulas Table of Contents

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: 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) Go to: Problem Formulas Table of Contents

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: 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 Go to: Problem Formulas Table of Contents