École de technologie supérieure MAT145 tutoring

1. Distinguish the quantities. Displacement is the signed change in position; distance is the total length of the path travelled, regardless of direction. Let $\Delta x$ denote displacement and $D$ total distance. Then $\Delta x=\int_0^5v(t)\,dt$, whereas $D=\int_0^5|v(t)|\,dt$. 2. Integrate the signed velocity. A convenient antiderivative is $F(t)=t^2/2-2t$, since $F'(t)=t-2=v(t)$; the prime denotes differentiation with respect to time. Therefore $\Delta x=F(5)-F(0)=(25/2-10)-0=2.5\,\mathrm{m}$. The positive answer means the final position is that far in the positive direction from the initial one. 3. Split where direction changes. The velocity is zero at $t=2$, negative before that time and positive after it. Thus $D=-\int_0^2(t-2)\,dt+\int_2^5(t-2)\,dt$. Since $F(2)=-2$, the first contribution is $-[F(2)-F(0)]=2\,\mathrm{m}$; the second is $F(5)-F(2)=2.5-(-2)=4.5\,\mathrm{m}$. Hence $D=6.5\,\mathrm{m}$. Taking $|\Delta x|$ would miss the earlier motion in the negative direction. 4. Check and interpret. The velocity graph forms triangles of area $\frac12(2)(2)=2$ and $\frac12(3)(3)=4.5$, confirming the two contributions. Changing the initial position shifts the entire position graph and the final position by the same amount. It changes neither the given velocity nor these integrals, so displacement and total distance remain unchanged.