Two students walk in the same direction along a straight path, at aconstant spee

Question

Two students walk in the same direction along a straight path, at aconstant speed - one at 0.90 m/s2 and the other at 1.90m/s2

a. Assuming that they start at the same point and thesame time, how much sooner does the faster student arrive at adestination 780 km away?
I used the formula: Average Velocity =?x/?t
1.90 m/s2 = 780 / t ... came out to 410seconds (my teacher wants me to round so that there aren'tdecimals)
0.90 m/s2 = 780/ t ... came out to 867seconds.
My problem is with part b. (But if I'm wrong on part a,please mention it..)

b. How far would the students have to walk so that the fasterstudent arrives 5.50 min before the slower student?
My first step was to convert 5.50 min to seconds - whichcomes out as 330 seconds. But after this, I'm stuck. Please helpand show all steps after this. Thanks in advance.

Explanation / Answer

d=780 km=780000 m
s=ut+1/2at2

t1=(2s/a1)=1316.56 s

t2=(2s/a2)=906.11 s

difference in their time=410 s

your question has a mistake

the data given are acceleration instead of velocity

because if velocity is constant the slower one will laways be behind

b)(2s/a2)-(2s/a1)=330

s*((2/0.9)-(2/1.9))=330

(s)=710.08

s=504219.54 m=504.219 km

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