# 2021 AMC 10A Problems/Problem 4

## Problem

A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$-second time interval, it travels $7$ inches more than during the previous $1$-second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

## Solution 1 (Arithmetic Series)

Since $$\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},$$ we seek the sum $$5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,$$ in which there are $30$ terms.

The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is $$5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.$$ 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: $$\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.$$ ~MRENTHUSIASM

## Solution 2 (Arithmetic Series)

The distance (in inches) traveled within each $1$-second interval is: $$5,5+1(7),5+2(7), \dots , 5+29(7).$$ This is an arithmetic sequence so the total distance travelled, found by summing them up is: $$\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.$$ Or, $$30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{\textbf{(D)} ~3195}.$$ ~BakedPotato66

## Solution 3 (Answer Choices and Modular Arithmetic)

From the $30$-term sum $$5+12+19+26+\cdots$$ in Solution 1, taking modulo $10$ gives $$5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.$$ The only answer choices congruent to $5$ modulo $10$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimation, $\textbf{(A)}$ is too small, leaving us with $\boxed{\textbf{(D)} ~3195}.$

~MRENTHUSIASM

## Solution 4 (Motion With Constant Acceleration)

This problem can be solved by physics method. The average speed increases $7 \ \text{in/s}$ per second. So, the acceleration $a=7 \ \text{in/s\textsuperscript{2}}.$ The average speed of the first second is $5 \ \text{in/s}.$ We can know the initial velocity $v_0=5-0.5\cdot7=1.5.$ The displacement at $t=30$ is $$s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.$$ ~Bran_Qin

~ pi_is_3.14

## Video Solution (Simple and Quick)

~ Education, the Study of Everything

## Video Solution (Arithmetic Sequence but in a Different Way)

~ North America Math Contest Go Go Go

~savannahsolver

~IceMatrix

## Video Solution by The Learning Royal

 2021 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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