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

Question

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

Infinity Learn · Accepted Answer

ν$$=$$α$$t1$$⇒$$t1=$$να$$v=$$β$$t2$$⇒$$t2=v$$β∴ $$t0=t-t1-t2=t-v$$α$$-v$$βNow, $$l=12$$α$$t12+vt0+12$$β$$t22=12($$α$$)v$$α$$2+vt-v$$α$$-v$$β$$+12($$β$$)v$$β$$2=vt-v22$$α$$-v22$$β∴ $$t=lv+v21$$α$$+1$$βFor t to be minimum its first derivation with respect to velocity be zero, i.$$e.0=-1v2+$$α$$+$$β$$2$$αβ∴$$v=2l$$αα$$+$$β

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

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