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Motion in One Dimension — PHYSICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9PHYSICSMotion in One Dimension3 Marks
Question
Figure given below shows a velocity-time graph for a car starting from rest. The graph has three parts $AB, BC$ and $CD.$

Compare the distance travelled in part BC with the distance travelled in part $AB$.
✓

Answer

Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
$= (2t - t) \times v_o$
$= v_ot$
Distance travelled in part AB = Area of the triangle ABt
$= (1/2) \times $ base × height
$= (1/2) \times t \times v_o$
$= (1/2) v_o t$
Therefore, distance travelled in part BC:distance travelled in part $AB :: 2:1.$

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