Two students walk in the same direction along a straight path at a constant speed. One walks at a speed of 0.90 m/s and the othe

Question

3 months ago

5

Two students walk in the same direction along a straight path at a constant speed. One walks at a speed of 0.90 m/s and the othe

r at a speed of 1.9 m/s. Assuming they have started at the same point at the same time, how much sooner does the faster student arrive at the destination 780 miles away?
A) 456 seconds
B) 411 seconds
C) 780 seconds
D) 867 seconds

Physics

2 answers:

Ostrovityanka [3.2K] · Answer 1 · 2026-03-07T01:41:47+03:00

Answer:

A. 456 seconds

Explanation:

Two students walk down a straight path in the same direction at constant velocities.

One has a velocity of 0.90 m/s

while the other moves at 1.9 m/s.

Each student covers a distance of 780 meters.

Our goal is to determine who is quicker and the additional time the slower student takes compared to the faster one.

For the student moving at 0.90 m/s, the time taken is $\frac{780}{0.90}$ .

Time= $\frac{distance}{speed}$ .

The time for the student at 0.90 m/s is

= $\frac{780}{0.90}$ .

This time calculates to 866.6 seconds.

Now for the student traveling at 1.9 m/s: their time is $\frac{780}{1.9}$ ,

which amounts to 410.5 seconds.

The second student, moving at 1.9 m/s, is faster than the one at 0.90 m/s.

The slower student takes 866.6 - 410.5 = 456.1 seconds longer.

Thus, the additional time for the slower student is 456 seconds.

Therefore, option A is correct.

Sav [3.1K] · Answer 2 · 2026-03-07T01:43:16+03:00

The correct answer is A) 456 seconds.

Further Explanation

Speed indicates the distance traveled over time. It is calculated using the distance and the time taken for that journey based on the following formula:

$speed = \frac{distance}{time}$ .

Firstly, we organize the provided information:

distance = 780 m

(NOTE: the distance in question is 780 meters, not miles. If it were miles, the options would be invalid.)

velocity of student 1 = 0.90 m/s

velocity of student 2 = 1.9 m/s

The query is seeking the time difference in arrival for both students. The procedure involves:

Calculating the time for each student to reach 780 m.
Finding out the time difference.

STEP 1: Using speed to compute the time to cover 780 meters, applying the general formula:

$time = \frac{distance}{speed}$ .

For Student 1:

$time = \frac{780 \ m}{0.90 \frac{m}{s}}\\\boxed {time = 866.67 \ s}$ .

For Student 2:

$time = \frac{780 \ m}{1.9 \frac{m}{s}}\\\boxed {time = 410.53 \ s}$ .

STEP 2: To find out how much sooner the faster student gets to the goal, we subtract their times.

$866.67 \ s \ - 410.53 \ s \ = \boxed {\boxed {\ 456.14 \ s}}$ .

This time difference indicates how long it takes before the slower student arrives.

The answer, rounded to the nearest whole number, confirms option A) 456 seconds.

Learn More

Learn more about velocity
Learn more about distance - time graphs
Learn more about distance

Keywords: speed, distance, time