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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle moves along a straight line with a variable acceleration...

A particle moves along a straight line with a variable acceleration given in the acceleration-displacement (a−S)\left(a - S\right) curve as shown in the figure. Determine the velocity (in ms−1m{s}^{- 1}) of the particle after it has travelled a distance of 30m30m. The initial velocity is 10ms−110\text{m}{s}^{- 1}.
A particle moves along a straight line with a variable acceleration given in the acceleration-displacement (a - S)

Correct answer

20

Step-by-step solution

A particle moves along a straight line with a variable acceleration given in the acceleration-displacement (a - S)

Area under curve is =12×10×30= \dfrac{1}{2} \times 10 \times 30
=150= 150 ……..(i)
Area under the curve is also equal to =v2−u22= \dfrac{{v}^{2} - {u}^{2}}{2} …….(ii)
From (i) and (ii)
12(v2−u2)=150\dfrac{1}{2}\left({v}^{2} - {u}^{2}\right) = 150
\left({v}^{2} - {u}^{2}\right) =300<br/><br />{v}^{2} = {u}^{2} + 300<br/><br />{v}^{2} = {\left(10\right)}^{2} + 300<br/><br />v = \sqrt{400} = 20\text{m}{s}^{- 1}
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