AP EAMCET · PHYSICS · Motion In One Dimension
The acceleration of a particle which moves along the positive \(x\)-axis varies with its position as shown in the figure. If the velocity of the particle is \(0.8 \mathrm{~ms}^{-1}\) at \(x=0\), then its velocity at \(x=1.4 \mathrm{m}\) is (in \(\mathrm{ms}^{-1}\))
- A 1.6
- B 1.2
- C 1.4
- D 0.8
Answer & Solution
Correct Answer
(B) 1.2
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {For a - } \mathrm{x} \text { graph } \\ & \mathrm{v}^2=\mathrm{u}^2+\int_{\mathrm{x}_1}^{\mathrm{x}_2} \mathrm{adx}=\mathrm{u}^2+2 \text { (Area of a-x graph) } \\ & \Rightarrow \mathrm{v}^2=(0.8)^2+2\left[0.4 \times 0.4+\frac{1}{2}(0.4+0.2) \times…
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