Question

The speed of train increases at a constant rate alfa from zero to V and then remains constant for an interval and finally decreses to zero at a constant rate Beta. If L be the total distance travelled then total time taken is?

Answer 1

(i) velocity inc. from 0 to V,
use first equation v = u + a*t : u = 0, v = V, a = α

this gives t = $\frac{V}{ \alpha }$

use second equation s = u*t + $\frac{1}{2}*a* t^{2}$ : u = 0, a = α, t = $\frac{V}{ \alpha }$

this gives s1 = $\frac{ V^{2} }{2* \alpha }$

(ii) velocity dec. from V to 0,
use first equation v = u + a*t : u = 0, v = V, a = -β

this gives t = $\frac{V}{ \beta }$

use second equation s = u*t + $\frac{1}{2}*a* t^{2}$ : u = V, a = -β, t = $\frac{V}{ \beta }$

this gives s2 = $\frac{ V^{2} }{2* \beta }$

So total distance traveled d = $\frac{ V^{2} }{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )$

(iii) Thus distance traveled during constant velocity phase = L - $\frac{ V^{2} }{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )$

So time taken to travel this distance T = $\frac{L - \frac{ V^{2} }{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )}{v}$ -> $\frac{L}{V} - \frac{V}{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )$

So total time taken to cover total distance L, = $(\frac{V}{ \alpha }) + (\frac{L}{V} - \frac{V}{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )) + (\frac{V}{ \beta })$
-> $\frac{L}{V} + \frac{V}{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )$

Answer
$\frac{L}{V} + \frac{V}{2} * ( \frac{1}{ \alpha } + \frac{1}{ \beta } )$

Answer 2

V = 0 + α t1      => t1 = V / α
S1 = 0 t + 1/2 α t1²  = 1/2 α V² / α² = 1/2 V²/α

S2 = V t2

0 = V - β t3    => t3 = V/β
S3 = (0+V)/2 * t3  = average speed * time
     =  1/2 V V / β = 1/2 V² / β
Total distance L = 1/2 V² (1/α + 1/β) + V t2
t2 =  [ L/V - 1/2 V(1/α+1/β) ]

Total time taken =   V/α + L /V - 1/2 V / α - 1/2 V/β + V/β
                   =  V/2 (1/α + 1/β ) + L/V