91Ó°ÊÓ

The problem is based on the simple concept of calculating area of triangle. The calculation of region bounded by the curve within the given graph is called as area under the curve.

Formula:

The area of triangle is given by,

$A=\frac{1}{2}\left(\mathrm{base}×\mathrm{height}\right)$

To find the coordinates at $5\sec$, we need to know the displacement of the particle at $5\sec$. The displacement of the particle would be equal to the area under the plot at $t=5\sec$.

Area of the triangle for$t=2\sec$is

$\begin{array}{c}{A}_1=\frac{1}{2}×2×4\\ =4\mathrm{m}\end{array}$

Area of the rectangle for$t=2\sec \mathrm{to}t=4\sec$is

$\begin{array}{c}{A}_2=2×4\\ =8\mathrm{m}\end{array}$

Area of the shape for$t=4\sec \mathrm{to}t=5\sec$

$\begin{array}{c}{A}_3=\left(\frac{1}{2}×1×2\right)+\left(1×2\right)\\ =3\mathrm{m}\end{array}$

So, the total displacement is

$\left(A_1+A_2+A_3\right)$

$4\mathrm{m}+8\mathrm{m}+3\mathrm{m}=15\mathrm{m}$

So, the coordinates at localid="1663063994425" $\mathrm{t}=5\sec$are$\left(5,15\right)$.

Velocity at $5.00\sec$is given from graph as $2\mathrm{m}/\mathrm{s}$.

Acceleration at $5.0\sec$can be found by finding the slope of the graph at $\mathrm{t}=5.0\sec$. It is equal to $\mathrm{a}=-2\mathrm{m}/{\mathrm{s}}^2$. It is negative since the velocity is reducing.

Using the method used in (a), we can find the displacement at $\mathrm{t}=1\sec$, which is equal to$1\mathrm{m}$, we already know the displacement at$\mathrm{t}=5\sec$which is$15\mathrm{m}$

$\begin{array}{c}{v}_{avg}=\frac{\mathrm{Δ}x}{\mathrm{Δ}t}\\ =\frac{15-1}{5-1}\\ =\frac{14}{4}\\ =3.5\mathrm{m}/\mathrm{s}\end{array}$

So, the average velocity would be$3.5\mathrm{m}/\mathrm{s}$.

The average acceleration at$1.0\sec$ and$5.0\sec$

Velocity at $t=1\sec$is$2\mathrm{m}/\mathrm{s}$and velocity at$t=5\sec$is$2\mathrm{m}/\mathrm{s}$

$\begin{array}{c}{\mathrm{a}}_{\mathrm{avg}}=\frac{\mathrm{Δ}v}{\mathrm{Δ}t}\\ =\frac{2-2}{4}\\ =0\end{array}$

So, the average acceleration would be$0$.