Question

The velocity of an object is changing with time and relation is given by the following equation. V=2t+3t^2 . calculate the position of the object from the orign at t=2s. Assume that the particle to be at origin at t=0.

Answer 1

V = $\frac{dx}{dt} = 2t +3t^{2}$

$dx =(2t+3t^{2})dt$

$dx = 2t.dt + 3t^{2} .dt$

$\int\limits dx = \int\limits 2t.dt +\int\limits 3t^{2}.dt$\

$x = \frac{2t^{2} }{2} + \frac{3t^{3} }{3}$               [ $\int\limits {n^{x} } \, dn  = \frac{n^{x+1} }{x+1}$ ]

now let us substitute t value [t=2]                    

$x = \frac{2(2)^{2} }{2} + \frac{3(2)^{3} }{3}$

x = 4 + 8

x  = 12 = distance

Answer 2

Answer:

Position of object is 12 m from the origin at t = 2s

Solution:

We know that,

v = $\dfrac{dx}{dt}$

Therefore, position x is given by :

x = $\displaystyle\int vdt$

$\longrightarrow$ x = $\displaystyle\int (2t + 3t^{2}) dt$

$\longrightarrow$ x = $\sf{\dfrac{2t^{2}}{2}}$ + $\sf{\dfrac{3t^{3}}{3}}$

$\longrightarrow$ x = t² + t³

Substitute t = 2s,

x = (2)² + (2)³

$\longrightarrow$ x = 4 + 8

$\longrightarrow$ x = 12 m

Hence, Position of object is 12 m from the origin at t = 2s