Question

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 other 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

Answer 1

Answer:

A. 456 seconds

Explanation:

We are given that two students walk in the same direction along a straight path at a constant speed.

One student walks with a speed=0.90 m/s

second student walks with speed=1.9 m/s

Total distance covered by each students=780 meter

We have to find who is faster and how much time  extra taken by slower student than the faster student.

Time taken by one student who travel with speed 0.90 m/s=$\frac{780}{0.90}$

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

Time taken by one student who travel with speed 0.90 m/s

=$\frac{780}{0.90}$

Time taken by one student who travel with speed 0.90 m/s

=866.6 seconds

Time taken by second student who travel with speed 1.9 m/s=$\frac{780}{1.9}$

=410.5 seconds

The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .

Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds

Extra time taken by the student travels with speed 0.90 m/s=456 seconds

Hence, option A is true.

Answer 2

The correct answer is A) 456 seconds.

Further Explanation

Speed is how far something travels per unit time. It is determined from the ratio of the distance covered and the time it took to cover such distance. It can be represented symbolically using the equation below:

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

To solve the problem, first sort the given:

distance traveled = 780 m

(NOTE: the destination is 780 meters from the origin not 780 miles away. None of the options will be viable if the distance was in miles.)

speed of student 1 = 0..90 m/s

speed of student 2 = 1.9 m/s

The question is asking to find the difference in the time of arrival of student 1 and student 2. To do this, the following steps must be done:

1. Calculate the time it will take for each student to arrive 780 m away from their origin.
2. Get the difference in the time.

STEP 1: Use the speed to calculate the time taken to travel a distance of 780 meters. The general equation to be used is:

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

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 determine how sooner the faster student reaches the destination, get the difference between their times.

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

The difference in time shows how many seconds elapse before the slower student arrives.

Rounding off the answer to the nearest whole number gives the correct answer which is A) 456 seconds.

Learn More

1. Learn more about velocity brainly.com/question/862972
2. Learn more about distance - time graphs brainly.com/question/1378025
3. Learn more about distance https://brainly.com/question/12971902

Keywords: speed, distance, time