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# Answer to Question #126301 in Mechanics | Relativity for jini

Question #126301
A train starting at station A moves along a straight line and stops at station B; distance x of the train from the starting point varies with time t as x= 3/4t^2- t^3/4
Find distance between station A and station B
i) 1unit ii)2 units iii) 1/2 unit
Find maximum velocity of the train.
i)3/4 unit ii)3/2 unit iii)1/4 unit

please show all the steps along with explanation and reply asap
1
Expert's answer
2020-07-16T10:58:45-0400

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}=0.5\\text{ 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}=1\\text{ 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}=0.16\\text{ 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."

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