A train travels along a straight horizontal track with constant acceleration. Points A, B and C are on the track with B between A and C. The distance AB is 1200 m and the distance BC is 2500 m. As the train passes B, its speed is 26ms–1. The train takes 60 s to travel from A to B.
c) Calculate the speed of the train as it passes C, giving your answer correct to one decimal place.
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The previous parts of this question are answered in;
It is given that the train travels 2500m from B to C.
Acceleration of the train is `0.2m/s^2` .
See the attached image with the calculations.
Let us say the velocity of C is Um/s and the time taken to travel between B and C as T seconds.
The acceleration from B to C is 0.2m/s^2.
`(U-26)/T = 0.2`
`U = 26+0.2T`
The area of the time velocity graph from B to C represents the distance travelled from B to C.
`2500 = (U+26)/2xxT`
`2500 = (26+0.2T+26)/2xxT`
`5000 = 52T+0.2T^2`
`0.2T^2+52T-5000 = 0`
`T = (-52+-sqrt((-52)^2-4xx0.2xx(-5000)))/(2xx0.2)`
`T = 74.7 `
`T = -334.7`
`T = 74.7` since it is greater than 0.
`U = 26+0.2T = 40.9`
So the speed of the train at C is 40.9m/s.
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