Question

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

Answer 1

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:

$$v'(t)=0=\frac{3}{2}-\frac{3t}2,\\ t=1.$$

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

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