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

a.Let the initial and final speeds of the first student be uand v therefore we get v2 - u2 = 2aS u = 0,a = 0.90 m/s2 and S = 780 km = 780 *103 m/s or v2 - 02 = 2 * 0.90 * 780 *103 or v2 = 1404 * 103 or v = 1184.9 m/s Let the time taken by the first student to reach thedestination be t therefore we get v = u + at or t = (v - u/a) = (1184.9 - 0/0.90) = 1316.6 s Let the initial and final speeds of the second student beu1 and v1 therefore we get v12 - u12 =2a1S u1 = 0 and a1 = 1.90 m/s2 or v12 - 02 = 2 * 1.90 * 780* 103 or v12 = 2964 * 103 or v1 = 1721.6 m/s Let the time taken by the second student to reach thedestination be t1 therefore we get v1 = u1 +a1t1 or t1 = (v1 - u1/a) = (1721.6- 0/1.90) = 906.1 s t = t - t1 = 1316.6 - 906.1 = 410.5 s The faster student arrives 410.5 s earlier than the otherstudent. b.The distance travelled by the faster student in 5.50 minis S = ut1 +(1/2)a1t12 u = 0,t1 = 5.50 min = 5.50 * 60 s = 330 s anda1 = 1.90 m/s2 or S = 0 * 330 + (1/2) * 1.90 * (330)2 = 103455 m =103.455 * 103 m = 103.455 km The total distance the students have to walk so that thefaster student arrives 5.50 min before the slower student is Stotal = 780 + 103.455 = 883.455 km or v12 - 02 = 2 * 1.90 * 780* 103 or v12 = 2964 * 103 or v1 = 1721.6 m/s Let the time taken by the second student to reach thedestination be t1 therefore we get v1 = u1 +a1t1 or t1 = (v1 - u1/a) = (1721.6- 0/1.90) = 906.1 s t = t - t1 = 1316.6 - 906.1 = 410.5 s The faster student arrives 410.5 s earlier than the otherstudent. b.The distance travelled by the faster student in 5.50 minis S = ut1 +(1/2)a1t12 u = 0,t1 = 5.50 min = 5.50 * 60 s = 330 s anda1 = 1.90 m/s2 or S = 0 * 330 + (1/2) * 1.90 * (330)2 = 103455 m =103.455 * 103 m = 103.455 km The total distance the students have to walk so that thefaster student arrives 5.50 min before the slower student is Stotal = 780 + 103.455 = 883.455 km or S = 0 * 330 + (1/2) * 1.90 * (330)2 = 103455 m =103.455 * 103 m = 103.455 km The total distance the students have to walk so that thefaster student arrives 5.50 min before the slower student is Stotal = 780 + 103.455 = 883.455 km

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