two student walk in the same direction along a straight path, at a constant speed one at 0.90m/s and the other at 1.90m/s assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780m away?

Question

two student walk in the same direction along a straight path, at a constant speed one at 0.90m/s and the other at 1.90m/s assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780m away?

Answer 1

To find out how much sooner the faster student arrives at the destination, we need to compare the time it takes for each student to travel the distance of 780 meters.

We can use the formula:

time = distance / speed

For the slower student:
time_slow = 780m / 0.90m/s ≈ 866.67 seconds

For the faster student:
time_fast = 780m / 1.90m/s ≈ 410.53 seconds

The faster student takes approximately 410.53 seconds to reach the destination, while the slower student takes approximately 866.67 seconds. Therefore, the faster student arrives sooner by:

time_difference = time_slow - time_fast
time_difference = 866.67 seconds - 410.53 seconds
time_difference ≈ 456.14 seconds

So, the faster student arrives around 456.14 seconds or 7.6 minutes sooner at the destination.

Answer 2

To calculate how much sooner the faster student arrives at the destination, we need to find the time it takes for both students to reach the destination.

Let's call the time it takes for the slower student to reach the destination t (in seconds). Since the slower student is moving at a constant speed of 0.90 m/s, we can use the formula distance = speed × time.

For the slower student:
distance = speed × time
780 m = 0.90 m/s × t
780 m / 0.90 m/s = t
t ≈ 866.67 seconds

Now, let's calculate the time it takes for the faster student to reach the destination. Using the same formula:

For the faster student:
distance = speed × time
780 m = 1.90 m/s × t'
780 m / 1.90 m/s = t'
t' ≈ 410.53 seconds

The faster student took approximately 410.53 seconds to reach the destination, while the slower student took approximately 866.67 seconds.

To find out how much sooner the faster student arrived, we subtract the time it took for the faster student from the time it took for the slower student:

Time difference = t - t'
Time difference ≈ 866.67 seconds - 410.53 seconds
Time difference ≈ 456.14 seconds

Therefore, the faster student arrives approximately 456.14 seconds sooner at the destination.