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Question: Displacement curve of a particle moving along a straight line is shown. Tangents at A and B make ang...

Displacement curve of a particle moving along a straight line is shown. Tangents at A and B make angles 450 and 1350 with positive x-axis respectively. Average acceleration of the particle between A and B is:

A

−2 m/s2

B

1 m/s2

C

– 1m/s2

D

Zero

Answer

−2 m/s2

Explanation

Solution

V→A=dx→Adt=1m/s{\overrightarrow{V}}_{A} = \frac{d{\overrightarrow{x}}_{A}}{dt} = 1m/s; V→B=dx→Bdt=−1m/s{\overrightarrow{V}}_{B} = \frac{d{\overrightarrow{x}}_{B}}{dt} = - 1m/s

a→av=V→B−V→AtB−tA=−1−11=−2m/s2{\overrightarrow{a}}_{av} = \frac{{\overrightarrow{V}}_{B} - {\overrightarrow{V}}_{A}}{t_{B} - t_{A}} = \frac{- 1 - 1}{1} = - 2m/s^{2}.