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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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 `

`Or `

`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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