A student is running on a straight track with a velocity of 5m per second. With help of a graph show how much distance is run by him in a minute.
Answers
Explanation:
For convenience, let the students (constant) speed be v0 and the buss initial position be x0. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero and the initial velocity of the bus is taken to be zero. The positions of student x1 and bus x2 as functions of time are then
x1=v0t,x2=x0+(l/2)at2
A. Setting x1=x2 and solving for the time t gives
t=(1/a)(v0±v02−2ax0)
= (1/0.2)((5.0)±52−2(0.2)(40.0)) = 10.40 s
The student will be likely to hop on the bus the first time she passes it [see part (d) for a discussion of the later time]. During this time, the student has run a distance v0t = (5 m/s)(10 s) = 50 m
B.The speed of the bus is v = at = 0.2 x 10 = 20 m/s
C.The results can be verified by noting that the x lines for the student and the bus intersect at two points.
D. At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up with her. At this later time, the buss velocity is (0.2 ms−2) (40 s) = 8.0 m/s.
E. No; v02<2ax0, and the roots of the quadratic are imaginary.When the student runs at 3.5 m/s the two lines do not intersect.
F. For the student to catch the bus, v02<2ax0 and so the minimum speed is 2(0.2ms−2(40m) = 4.0 m/s. She would be running for a time 0.2ms−24.0m/s = 20s, and cover a distance of (4.0 m/s)(20 s) = 80.0 m.
However, when the student runs at 4.0 m/s, the lines intersect at one point (x = 80 m).
Explanation:
problem is solved in picture
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