Answer in Mechanics | Relativity for jini #126301

Find distance between station A and station B for the given units of time:

i) If it takes 1 unit of time to travel from A to B, the distance is

“x(1)=frac{3}{4}cdot1^2-frac{1^3}{4}=0.5text{ units of distance}.”

ii) If it takes 2 units of time to travel from A to B, the distance is

“x(2)=frac{3}{4}cdot2^2-frac{2^3}{4}=1text{ unit of distance}.”

ii) If it takes 1/2 units of time to travel from A to B, the distance is

“x(0.5)=frac{3}{4}cdot0.5^2-frac{0.5^3}{4}=0.16text{ units of distance}.”

To find the maximum velocity of the train, determine the equation for the velocity from the equation of displacements:

“v(t)=x'(t)=frac{3}{2}t-frac{3t^2}{4}.”

To find at what moment the velocity was maximum, equate the derivative of velocity to zero and find time:

“v'(t)=0=frac{3}{2}-frac{3t}2,\nt=1.”

However, we are given smaller values. For these cases, the velocity was

i) 3/4 unit

“v(3/4)=frac{3}{2}(3/4)-frac{3(3/4)^2}{4}=-0.21.”

ii) 3/2 unit

“v(3/2)=frac{3}{2}(3/2)-frac{3(3/2)^2}{4}=1.9.”

iii) 1/4 unit

“v(1/4)=frac{3}{2}(1/4)-frac{3(1/4)^2}{4}=-12.”