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Motion in One Dimension — PHYSICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9PHYSICSMotion in One Dimension2 Marks
Question
From the diagram given below, calculate
1. acceleration
2. deceleration
3. distance covered by body.
Image
(i) Acceleration $=$ Slope of $v$-t graph from O to A
$
=\frac{AC}{OC}=\frac{15}{10}=1.5 ms^{-2}
$
✓

Answer

(ii) Deceleration $=$ Slope of $v$. .t. graph from A to B
$
=\frac{A C}{C B}=\frac{15}{16-10}=\frac{15}{6}=2.5 ms^{-2}
$
(iii) Distance covered $=$ area under $v$.t. graph $=$
area of $\triangle OAC +$ area of $\triangle ABC$ $=\frac{1}{2} \times OC \times AC +\frac{1}{2} \times BC \times AC$ $=\frac{1}{2} \times 10 \times 15+\frac{1}{2} \times 6 \times 15$
$
\begin{array}{l}
=5 \times 15+3 \times 15 \\
=75+45=120 m
\end{array}$

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