A subway train starting from rest leaves a station with a constant acceleration. At the end of 5.59 s, its is moving at 9.9502 m/s. What is the train's displacement in the first 4.26517 s of motion? Answer in units of m.
When I used the formula d = v1t + 1/2at^2 I got 27.81 m which turns out to be the wrong answer. Please help!
2007-09-23 06:08:49 · 2 answers · asked by jon s 2 in Science & Mathematics ➔ Physics
First, the train's acceleration is:
a = 9.9502 (m/s)/5.59 (s) = 1.78(m/s^2)
So the required train's displacement in the first 4.26517 s of motion is:
0.5*a*t^2 = 0.5*1.78(m/s^2)* (4.26517s)^2
= 16.2m
2007-09-24 17:29:35 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋
we could mark east as +x direction and west as -x direction. initially driving force is on the beginning place. utilising this convention, first he strikes a vector +8km and then -3 km vector and ultimately +12 km then addition of those vectors Vector R= vectorx1 + vector2 + vectorx3 Vector R=(+8) + (-3) + (+12) Vector R=+17 we could convert it to vector lower back +x east direction so he strikes 17 km in east direction as displacement take care
2016-10-05 05:41:49 · answer #2 · answered by ? 4 · 0⤊ 0⤋