Question

a bus moves away from a bus stop in a straight line. it moves 10 meters by the end of the 1st second, 20 meters by the end of the 2nd second, and 30 meters by the end of the 3rd second. how many meters away from the bus stop after 10 seconds?
a. 100 c. 120
b. 110 d. 130
please select the best answer from the choices provided
a
b
c
d

Explanation:

Step1: Identify the pattern

The bus moves 10 meters in the 1st second (end of 1st second: 10m), 20 meters by end of 2nd second (10 + 10? Wait, no—wait, the problem says "moves 10 meters by the end of the 1st second, 20 meters by the end of the 2nd second, 30 meters by the end of the 3rd second". Wait, maybe it's a linear pattern: distance \( d = 10t \), where \( t \) is time in seconds. Let's check: at \( t = 1 \), \( d = 10(1) = 10 \); \( t = 2 \), \( d = 20 \); \( t = 3 \), \( d = 30 \). Yes, that's a linear relationship with slope 10 (constant speed of 10 m/s? Wait, no—wait, the distance at end of \( t \) seconds is \( 10t \). Wait, but maybe the problem has a typo, but assuming the pattern is 10m per second (since 10, 20, 30 at 1,2,3 seconds). Then at \( t = 10 \), distance \( d = 10 \times 10 = 100 \)? Wait, no—wait, the options include 100 (a), 110 (b), 120 (c), 130 (d). Wait, maybe I misread. Wait, maybe the bus moves 10m in first second, then 10m more each second? Wait, no—wait, the problem says "moves 10 meters by the end of the 1st second, 20 meters by the end of the 2nd second, 30 meters by the end of the 3rd second". So the distance at time \( t \) (in seconds) is \( 10t \). So at \( t = 10 \), \( 10 \times 10 = 100 \). But wait, maybe the problem is that the bus's distance is \( 10t \), so after 10 seconds, 1010=100. But let's check the options: option a is 100. Wait, but maybe I made a mistake. Wait, maybe the problem is that the bus accelerates? No, the distances are 10,20,30—linear, so constant speed of 10 m/s. So distance at \( t \) seconds is \( 10t \). So at \( t = 10 \), 1010=100. So the answer should be a. 100. Wait, but let's confirm. If at end of 1st second: 10m, end of 2nd: 20m (so 10m per second), end of 3rd: 30m. So yes, linear. So formula \( d = 10t \). So \( t = 10 \), \( d = 10*10 = 100 \). So option a.

Step2: Calculate distance at t=10

Using the pattern \( d = 10t \), substitute \( t = 10 \): \( d = 10 \times 10 = 100 \).

Answer:

A. 100