Question

If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/$s^2$ and subway stations are located 880 m apart. What is the maximum speed, a subway train can attain between stations?

• A. 38 m/s
• B. 68.6 m/s
• C. 34.3 m/s
• D. 51.2 m/s

Answer 1

The correct option is C

34.3 m/s

The acceleration time curve for the given situation will be

Now write equation

a = 1.34 m/$s^2$

The distance between 2 railway stops is 880 m.

So, the train travels with acceleration 1.34 m/$s^2$ fot the half journey i.e.440 m

If the total time taken for the journey is T then the train is accelerated for T/2 sec

a = $\frac{dv}{dt}$ = 1.34

${\int }_0^{v}dv={\int }_0^{t}1.34dt$

v = 1.34 t ....... (1)

for the time $\frac{T}{2}$ displacement is 440 m

Now v = $\frac{ds}{dt}$ = 1.34t

= $\Rightarrow {\int }_0^{440}ds={\int }_0^{\frac{T}{2}}$ 1.34t dt

440 = 1.34 $\frac{t^2}{2}{]}_0^{\frac{T}{2}}$

$\Rightarrow$ T = 51.25 sec.

Now substitute $\left(\frac{T}{2}\right)$ in equation (1)

V = 1.34 × $\left(\frac{51.25}{2}\right)$

V = 34.335 m/s