Question

The position of a particle along x axis at time t is given by :
x= 2 + t - 3t²
Then find out the displacement and the distance travelled in the interval t=0 to t=1 .

Answer 1

x = 2 +t -3t² ---------(1)

differentiate wrt time

dx/dt = 1 -6t ----------(2)

again differentiate wrt time

d²x/dt² = -6

hence, motion is retarding becoz d²x/dt² < 0

let at t time velocity of particle is zero

e.g dx/dt = 0 = 1-6t = 1/6

at t = 1/6 velocity of particle is zero .
then particle moves in back direction at t = 1 sec

now,
displacement = final position - initial position = x(1) - x (0)
= 2 + 1 -3 - 2
= -2 m

distance = | x(1) - x (1/6) | + | x (1/6) - x(0) |
= | 2 + 1 -3 -(2 +1/6 -3/36 | + | 2 + 1/6 -3/36 - 2 |

=| -2 - 1/6 +1/12 | + | 2 +1/6 -1/12 -2 |

= | -2 -1/12 | + | 1 /12 |

= 2 + 1/12 + 1/12

= 2 + 1/6 = 13/6 m

Answer 2

this is not correct answer you are moulding your answer for getting the given result insteed the correct answer is 2.2