Question

The speed of a train increases at a constant rate `alpha` from zero to v and then remains constant for an interval and finally decreases to zero at a constant rate `beta`. The total distance travelled by the train is l. The time taken to complete the journey is t. Then,

Answer 1

`A` to `B`: `v=alphat_(1)impliest_(1)=v/alpha`
`v^(2)=2 alphas_(1)impliess_(1)=v^(2)/(2alpha)`
`B` to `C`: `s_(2)=vt_(2)impliest_(2)=s_(2)/v`
`C` to `D`: `0=v-betat_(3)impliest_(3)=v/beta`
`0=v^(2)-2betas_(3)impliess_(3)=v^(2)/(2beta)`
`t_(2)=s_(2)/v=(d-(s_(1)+s_(3)))/v`
`=d/v-(s_(1)/v+s_(3)/v)`
`=b/v-(v)/(2alpha)+v/(2beta){Since s_(1)=v^(2)/(2alpha),s_(3)=v^(2)/(2beta)}`
Total time `t=t_(1)+t_(2)+t_(3)`
`=v/alpha+d/v-(v/(2alpha)+v/(2beta))+v/beta`
`=d/v+v/2(1/alpha+1/beta)`