Two students walk in the same direction along a straight path, at a constant speed—one at 0.90 m/s and the other at 1.90 m/s. a. assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away? b. how far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

a)456.15s b)564.3m

Step-by-step explanation:

Let be student 1 the one with V1=0.90m/s as his speed, T1 his time and X1 his distance.

Let be then student 2 the one with V2=1.90m/s as his speed, T2 his time and X2 his distance.

For both

because of their constant speed.

then,

and

then the second students will arrive 866.67-410.52=456.15 seconds sooner.

to get 5.50min =60s . 5.50=330 seconds sooner T2+330s=T1 and X1=X2

they have to make 564.3m so the faster kid gets 5.50 minutes sooner

its 50 miles away from the starting point