Question

A particle moves along x-axis is velocity is a function of time according to relation V =(3t²-18t +24) m/s assume that at t = 0 ,particle is at orgin then the distance travelled by the particle in 0 to 3 seconds in time interval is

Answer 1

Answer:

V= ( 3t ^2 - 18t + 24 )

differentiate ---------

v = 6t- 18

The distance travelled by the particle in 3 second --------

= 6×3-18

=18 -18

=0 Ans.

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Answer 2

Answer:

they have given the velocity function,which on differntiantion will give acceleration but we don't want acceleration. For finding distance integrate the velocity function, i.e; on integrating

V=($3t^{2} -18t+24$) we get  S=($t^3-9t^2+24t$)

now distance travelled in 0 to 3 seconds will be

$\int\limits^3_0 {t^3-9t^2+24t} \, dt\\ \\ \int\limits^3_0 {t^3}dt - 9\int\limits^3_0 {t^2}dt + 24\int\limits^3_0 {t}dt$

$S=(3-0)^3-9(3-0)^2+24(3-0)$

∴ S=18m

Explanation: