Question
✓
Answer
(i) Velocity remains the same from 0 to 2 sec .
$\therefore$ Acceleration will be zero at time $t =1 sec$.
(ii) Velocity at $2 sec 5 m / s$ and the velocity at 4 sec in hence the acceleration between $2-4 sec$.
$
\begin{array}{l}
=\frac{0-5}{4-2} \\
=\frac{-5}{2}=-2.5 m / s^2Ans.
\end{array}
$
(iii) Distance $=$ Area of rectangle GCAF + Area of triangle CBA + Area of triangle BED
$
\begin{array}{l}
=5 \times 2+\frac{1}{2}(4-2) \times 5+\frac{1}{2}(6-4) \times 5 \\
=10+5+5=20 \text { meter }Ans.
\end{array}
$
Displacement $=$ Area of rectangle GCAE + Area of triangle CBA + Area of triangle BED
$
\begin{array}{l}
=5 \times 2+\frac{1}{2}(4-2) \times 5-\frac{1}{2}(6-4) \times 5 \\
=10+5-5=10 \text { meter }Ans.
\end{array}$
$\therefore$ Acceleration will be zero at time $t =1 sec$.
(ii) Velocity at $2 sec 5 m / s$ and the velocity at 4 sec in hence the acceleration between $2-4 sec$.
$
\begin{array}{l}
=\frac{0-5}{4-2} \\
=\frac{-5}{2}=-2.5 m / s^2Ans.
\end{array}
$
(iii) Distance $=$ Area of rectangle GCAF + Area of triangle CBA + Area of triangle BED
$
\begin{array}{l}
=5 \times 2+\frac{1}{2}(4-2) \times 5+\frac{1}{2}(6-4) \times 5 \\
=10+5+5=20 \text { meter }Ans.
\end{array}
$
Displacement $=$ Area of rectangle GCAE + Area of triangle CBA + Area of triangle BED
$
\begin{array}{l}
=5 \times 2+\frac{1}{2}(4-2) \times 5-\frac{1}{2}(6-4) \times 5 \\
=10+5-5=10 \text { meter }Ans.
\end{array}$
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