Question

The position of a particle moving along a straight line is defined by the relation, x = t³ – 6t² – 15 t + 40 where x is in meters and t in seconds. The distance travelled by the particle from t = 0 to t = 2 s is

	44 m
	112 m
	46 m
	54 m

Answer 1

Dear student,
$$\begin{gathered} positionoftheparticleisgivenas, \\ x=t^3–6t^2–15t+40 \\ attimet=0 \\ {\overline{)x}}_{t=0}=+40 \\ attimet=2 \\ {\overline{)x}}_{t=2}=2^3-6\times 2^2-15\times 2+40=-6 \\ dis\tan cetravelled={\overline{)x}}_{t=2}-{\overline{)x}}_{t=0}=-6-40=-46m \\ henceoption3 \\ -vesignonlydepictstahttheparticleismovingalong-vexaxis. \end{gathered}$$

Regards,