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

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=

Time=

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

=

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=

=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.

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever