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==Solution 4 (Motion With Constant Acceleration)== | ==Solution 4 (Motion With Constant Acceleration)== | ||
| − | This problem can be solved by physics method. The average speed increases <math>7 \ \text{in/s}</math> per second. So, the acceleration <math>a=7 \ \text{in/s\textsuperscript{2}}.</math> The average speed of the first second is <math>5 \ \text{in/s}.</math> We can know the initial velocity <math>v_0=5-0.5\cdot7=1.5.</math> The displacement at <math>t=30</math> is <cmath>s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.</cmath> | + | This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases <math>7 \ \text{in/s}</math> per second. So, the acceleration <math>a=7 \ \text{in/s\textsuperscript{2}}.</math> The average speed of the first second is <math>5 \ \text{in/s}.</math> We can know the initial velocity <math>v_0=5-0.5\cdot7=1.5.</math> The displacement at <math>t=30</math> is <cmath>s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.</cmath> |
~Bran_Qin | ~Bran_Qin | ||
Latest revision as of 06:59, 13 November 2023
Contents
- 1 Problem
- 2 Solution 1 (Arithmetic Series)
- 3 Solution 2 (Arithmetic Series)
- 4 Solution 3 (Answer Choices and Modular Arithmetic)
- 5 Solution 4 (Motion With Constant Acceleration)
- 6 Video Solution by OmegaLearn
- 7 Video Solution (Simple and Quick)
- 8 Video Solution (Arithmetic Sequence but in a Different Way)
- 9 Video Solution
- 10 Video Solution by TheBeautyofMath
- 11 Video Solution by The Learning Royal
- 12 See Also
Problem
A cart rolls down a hill, travelling 
inches the first second and accelerating so that during each successive 
-second time interval, it travels 
inches more than during the previous -second interval. The cart takes 
seconds to reach the bottom of the hill. How far, in inches, does it travel?
Solution 1 (Arithmetic Series)
Since ![\[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\]](http://latex.artofproblemsolving.com/3/c/c/3cc286b6c23404aef94cb7500cd7f604bb3cf900.png)
we seek the sum ![\[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\]](http://latex.artofproblemsolving.com/a/6/5/a65d35f502f5a93a00e576aeb6d1b99c8af2464c.png)
in which there are terms.
The last term is 
Therefore, the requested sum is ![\[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.\]](http://latex.artofproblemsolving.com/4/f/5/4f5f50ede4de794fe97e1299c26b81ffa228c402.png)
Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms: ![\[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\]](http://latex.artofproblemsolving.com/c/e/6/ce65ba615dc4e73740078ca1f10a5fb4189d7a9c.png)
~MRENTHUSIASM
Solution 2 (Arithmetic Series)
The distance (in inches) traveled within each -second interval is: ![\[5,5+1(7),5+2(7), \dots , 5+29(7).\]](http://latex.artofproblemsolving.com/e/4/4/e44d5fff397c71a03c7d42feea02a054f9707604.png)
This is an arithmetic sequence so the total distance travelled, found by summing them up is:
![\[\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.\]](http://latex.artofproblemsolving.com/5/a/a/5aa6041c6f333f5fec5761bb314e49b07f806d51.png)
Or, ![\[30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{\textbf{(D)} ~3195}.\]](http://latex.artofproblemsolving.com/b/7/9/b7960a15d128e6c021c8702eb9bd0360a29e839e.png)
~BakedPotato66
Solution 3 (Answer Choices and Modular Arithmetic)
From the -term sum ![\[5+12+19+26+\cdots\]](http://latex.artofproblemsolving.com/b/d/f/bdf805a7ab90ac77957f3afc6261f69f0b3a72c3.png)
in Solution 1, taking modulo 
gives ![\[5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.\]](http://latex.artofproblemsolving.com/1/5/7/157e655192b7b968e33f4926b89d91540b39f00f.png)
The only answer choices congruent to modulo are 
and 
By a quick estimation, is too small, leaving us with 
~MRENTHUSIASM
Solution 4 (Motion With Constant Acceleration)
This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases 
per second. So, the acceleration 
The average speed of the first second is 
We can know the initial velocity 
The displacement at 
is ![\[s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.\]](http://latex.artofproblemsolving.com/4/8/b/48be2ca0210a2bc9ae92efff246f59d16f1e9111.png)
~Bran_Qin
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution (Simple and Quick)
~ Education, the Study of Everything
Video Solution (Arithmetic Sequence but in a Different Way)
https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4
~ North America Math Contest Go Go Go
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM?t=262
~IceMatrix
Video Solution by The Learning Royal
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
